inequality

Lê Anh Duy

03/05/2019 at 12:58

Votes Answers Follow

35*) Given $\Delta ABC$. Prove that: $S_{ABC}\le\dfrac{\sqrt{3}}{12}\left(AB^2+BC^2+CA^2\right)$

inequality Geometry area

Lê Anh Duy

27/02/2019 at 09:55

Votes Answers Follow

30) Let a, b, c are the length of a triangle. Prove that:

$\dfrac{ab}{a+b-c}+\dfrac{bc}{b+c-a}+\dfrac{ca}{c+a-b}\ge a+b+c$

inequality

FacuFeri 28/02/2019 at 15:58

Put $A=\dfrac{ab}{a+b-c}+\dfrac{bc}{b+c-a}+\dfrac{ac}{c+a-b}$

Because a ; b ; c are the length of a triangle so $a+b-c;b+c-a;c+a-b>0$

Put $a+b-c=x;b+c-a=y;c+a-b=z$

$\Rightarrow\dfrac{x+y}{2}=b;\dfrac{y+z}{2}=c;\dfrac{x+z}{2}=a;a+b+c=x+y+z$

We have : $\dfrac{\left(x+y\right)\left(x+z\right)}{2.2x}+\dfrac{\left(x+y\right)\left(y+z\right)}{2.2y}+\dfrac{\left(x+z\right)\left(y+z\right)}{2.2z}$

$=\dfrac{x\left(x+y+z\right)+yz}{4x}+\dfrac{y\left(x+y+z\right)+xz}{4y}+\dfrac{z\left(x+y+z\right)+xy}{4z}$

$=\dfrac{x+y+z}{4}+\dfrac{x+y+z}{4}+\dfrac{x+y+z}{4}+\dfrac{yz}{4x}+\dfrac{xz}{4y}+\dfrac{xy}{4z}$

$=\dfrac{3\left(x+y+z\right)}{4}+\dfrac{y^2z^2}{4xyz}+\dfrac{x^2z^2}{4xyz}+\dfrac{x^2y^2}{4xyz}$

Applying the inequality $a^2+b^2+c^2\ge ab+bc+ac$ , we have :

$A\ge\dfrac{3\left(x+y+z\right)}{4}+\dfrac{xyz\left(x+y+z\right)}{4xyz}=x+y+z=a+b+c$

Equal sign occurs $\Leftrightarrow x=y=z\Leftrightarrow a=b=c$
```
 
```
Lê Anh Duy selected this answer.
Nguyễn Thị Linh 06/03/2019 at 23:34

FacuFeri copy internet

Nguyễn Nhật Minh

02/04/2017 at 22:06

Votes Answers Follow

Prove that $\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\left(n\in N;n\ge2\right)$

Help me please!

Fraction inequality

Nguyễn Huy Tú 03/04/2017 at 13:03

$\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+2\right)}$

$=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{\left(n-1\right)n\left(n+2\right)}\right)$

$=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+2\right)}\right)$

$=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)}\right)$

$=\dfrac{1}{4}.\dfrac{1}{2n\left(n+2\right)}$

$\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n\left(n+2\right)}< \dfrac{1}{4}$

$\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}$

So $\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}$
Nguyễn Nhật Minh selected this answer.
AI kết bạn với mình là may mắn cả đời 04/04/2017 at 21:05

$ \dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+2\right)}$ $=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{\left(n-1\right)n\left(n+2\right)}\right)$ $=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+2\right)}\right)$ $=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)}\right)$ $=\dfrac{1}{4}.\dfrac{1}{2n\left(n+2\right)}$ $\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n\left(n+2\right)}< \dfrac{1}{4}$ $\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}$ So $\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}$ $\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+2\right)}$ $=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{\left(n-1\right)n\left(n+2\right)}\right)$ $=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+2\right)}\right)$ $=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)}\right)$ $=\dfrac{1}{4}.\dfrac{1}{2n\left(n+2\right)}$ $\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n\left(n+2\right)}< \dfrac{1}{4}$ $\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}$ So $\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}$ $\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+2\right)}$ $=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{\left(n-1\right)n\left(n+2\right)}\right)$ $=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+2\right)}\right)$ $=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)}\right)$ $=\dfrac{1}{4}.\dfrac{1}{2n\left(n+2\right)}$ $\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n\left(n+2\right)}< \dfrac{1}{4}$ $\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}$ So $\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}$
Indrace AD-MIN 07/04/2017 at 21:35

123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14 123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14 123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14

Nguyễn Nhật Minh

04/04/2017 at 06:29

Votes Answers Follow

Prove that:

$\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}< 1$(n $\in$N*)

Fraction inequality

¤« 19/02/2018 at 21:29

$\frac{3}{4} + \frac{5}{36} + \frac{7}{144} + . . . + \frac{2 n + 1}{n^{2} {(n + 1)}^{2}}$

$= 1 - \frac{1}{2^{2}} + \frac{1}{2^{2}} - \frac{1}{3^{2}} + \frac{1}{3^{2}} - \frac{1}{4^{2}} + . . . + \frac{1}{{(n - 1)}^{2}} + \frac{1}{n^{2}}$

$= 1 - \frac{1}{n^{2}} < 1$
Lê Quốc Trần Anh Coodinator 12/02/2018 at 09:27

$\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}$

$=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{\left(n-1\right)^2}+\dfrac{1}{n^2}$

$=1-\dfrac{1}{n^2}< 1$
FC Alan Walker 21/02/2018 at 17:38

Ta có: $\dfrac{2n+1}{n^2\left(n+1\right)^2}=\dfrac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}=\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}$

Do đó $\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}$

$=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}$

$=1-\dfrac{1}{\left(n+1\right)^2}< 1$

Vậy $\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}< 1\forall n\in$N*

Macedoi

07/08/2017 at 09:26

Votes Answers Follow

Given $a,b,c$ are non-negative numbers such that $ab+bc+ca=1$. Find the minimize value of $P=\dfrac{1}{\sqrt{a^2+b^2}}+\dfrac{1}{\sqrt{b^2+c^2}}+\dfrac{1}{\sqrt{c^2+a^2}}$

-Source: Câu hỏi của michelle holder - Toán lớp 10 | Học trực tuyến (Ace Legona's solution is wrong)

inequality mixing-variable maxima-minima

FA KAKALOTS 28/01/2018 at 22:06

i have a solution but it's ugly

If $a = b = 1$

and $c = 0$ then we get a value $2 + \frac{1}{\sqrt{2}}$

We'll prove that it's a minimal value. Thus, we need to prove that

$\sqrt{\frac{a b + b c + c a}{a^{2} + b^{2}}} + \sqrt{\frac{a b + b c + c a}{b^{2} + c^{2}}} + \sqrt{\frac{a b + b c + c a}{a^{2} + c^{2}}} \geq 2 + \frac{1}{\sqrt{2}}$

WLOG $c = min {a, b, c}$

. Hence:

$\frac{a b + a c + b c}{a^{2} + b^{2}} - \frac{(a + c) (b + c)}{(a + c)^{2} + (b + c)^{2}} = \frac{c (a + b + 2 c) (2 a b + a c + b c)}{a^{2} + b^{2}) ((a + c)^{2} + (b + c)^{2}} \geq 0$

Similar $\frac{a b + a c + b c}{a^{2} + c^{2}} - \frac{b + c}{a + c} = \frac{c (2 a b + a c - c^{2})}{(a + c) (a^{2} + c^{2})} \geq 0$

And $\frac{a b + a c + b c}{b^{2} + c^{2}} - \frac{a + c}{b + c} = \frac{c (2 a b + b c - c^{2})}{(b + c) (b^{2} + c^{2})} \geq 0$

Let $\frac{a + c}{b + c} = x^{2}; \frac{b + c}{a + c} = y^{2} (x, y > 0)$

$\Rightarrow x y = 1$

and we have:

$x + y + \frac{1}{\sqrt{x^{2} + y^{2}}} \geq 2 + \frac{1}{\sqrt{2}}$

$\Leftrightarrow x + y - 2 \sqrt{x y} \geq \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{x^{2} + y^{2}}}$

$\Leftrightarrow (\sqrt{x} - \sqrt{y})^{2} \geq \frac{(x - y)^{2}}{\sqrt{2 (x^{2} + y^{2})} (\sqrt{x^{2} + y^{2}} + \sqrt{2})}$

$\Leftrightarrow \sqrt{2 (x^{2} + y^{2})} (\sqrt{x^{2} + y^{2}} + \sqrt{2}) \geq (\sqrt{x} + \sqrt{y})^{2}$

By Cauchy-Schwarz's ine we have:

$\sqrt{2 (x^{2} + y^{2})} = \sqrt{(1^{2} + 1^{2}) (x^{2} + y^{2})} \geq x + y$

$= \frac{1}{2} (1^{2} + 1^{2}) ((\sqrt{x})^{2} + (\sqrt{y})^{2}) \geq \frac{1}{2} (\sqrt{x} + \sqrt{y})^{2}$

Thus, it's enough to prove that $\sqrt{x^{2} + y^{2}} + \sqrt{2} \geq 2$

It's true by AM-GM $\sqrt{x^{2} + y^{2}} + \sqrt{2} \geq \sqrt{2 x y} + \sqrt{2} = 2 \sqrt{2} > 2$
AL 08/08/2017 at 13:15

i have a solution but it's ugly

If $a=b=1$ and $c=0$ then we get a value $2+\frac{1}{\sqrt2}$

We'll prove that it's a minimal value. Thus, we need to prove that

$\sqrt{\dfrac{ab+bc+ca}{a^2+b^2}}+\sqrt{\dfrac{ab+bc+ca}{b^2+c^2}}+\sqrt{\dfrac{ab+bc+ca}{a^2+c^2}}\ge2+\dfrac{1}{\sqrt{2}}$

WLOG $c=\min\{a,b,c\}$. Hence:

$\dfrac{ab+ac+bc}{a^2+b^2}-\dfrac{(a+c)(b+c)}{(a+c)^2+(b+c)^2}=\dfrac{c(a+b+2c)(2ab+ac+bc)}{a^2+b^2)((a+c)^2+(b+c)^2}\ge0$

Similar$\dfrac{ab+ac+bc}{a^2+c^2}-\dfrac{b+c}{a+c}=\dfrac{c(2ab+ac-c^2)}{(a+c)(a^2+c^2)}\ge0$

And $\dfrac{ab+ac+bc}{b^2+c^2}-\dfrac{a+c}{b+c}=\dfrac{c(2ab+bc-c^2)}{(b+c)(b^2+c^2)}\ge0$

Let $\dfrac{a+c}{b+c}=x^2;\dfrac{b+c}{a+c}=y^2\left(x,y>0\right)$$\Rightarrow xy=1$ and we have:

$x+y+\dfrac{1}{\sqrt{x^2+y^2}}\ge2+\dfrac{1}{\sqrt{2}}$

$\Leftrightarrow x+y-2\sqrt{xy}\ge\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{x^2+y^2}}$

$\Leftrightarrow(\sqrt{x}-\sqrt{y})^2\ge\dfrac{(x-y)^2}{\sqrt{2(x^2+y^2)}(\sqrt{x^2+y^2}+\sqrt{2})}$

$\Leftrightarrow\sqrt{2(x^2+y^2)}(\sqrt{x^2+y^2}+\sqrt{2})\ge(\sqrt{x}+\sqrt{y})^2$

By Cauchy-Schwarz's ine we have:

$\sqrt{2(x^2+y^2)}=\sqrt{(1^2+1^2)(x^2+y^2)}\ge x+y$

$=\dfrac{1}{2}(1^2+1^2)((\sqrt{x})^2+(\sqrt{y})^2)\ge\dfrac{1}{2}(\sqrt{x}+\sqrt{y})^2$

Thus, it's enough to prove that $\sqrt{x^2+y^2}+\sqrt{2}\ge2$

It's true by AM-GM $\sqrt{x^2+y^2}+\sqrt{2}\ge\sqrt{2xy}+\sqrt{2}=2\sqrt{2}>2$
Phan Huy Toàn 20/08/2017 at 20:10

tôi để lại số quả táo là:

Aim Egst

25/07/2017 at 19:04

Votes Answers Follow

Given $m\ge n>0$ and $a,b,c$ are positive real numbers. Prove this inequality $\dfrac{a^m}{b^n+c^n}+\dfrac{b^m}{c^n+a^n}+\dfrac{c^m}{a^n+b^n}\ge\dfrac{a^{m-n}+b^{m-n}+c^{m-n}}{2}$

it's a general formula of Nesbitt's inequality, i have a method but i need more :)), Help me, thanks

inequality

FA KAKALOTS 28/01/2018 at 22:06

My try: WLOG a≥b≥c hence we have: am−n≥bm−n≥cm−n

⇒anbn+cn≥bncn+an≥cnan+bn

By Chebyshev's inequality we have:

∑ambn+cn=∑am−n⋅anbn+cn

≥am−n+bm−n+cm−n3⋅∑anbn+cn

≥∑am−n+bm−n+cm−n2

We're done when a=b=c

P/s: This's discuss not A4 (Auto Ask Auto Answer)
Aim Egst 25/07/2017 at 19:12

My try: WLOG $a\ge b\ge c$ hence we have: $a^{m-n}\ge b^{m-n}\ge c^{m-n}$

$\Rightarrow\dfrac{a^n}{b^n+c^n}\ge\dfrac{b^n}{c^n+a^n}\ge\dfrac{c^n}{a^n+b^n}$

By Chebyshev's inequality we have:

$\text{∑}\dfrac{a^m}{b^n+c^n}=\text{∑}a^{m-n}\cdot\dfrac{a^n}{b^n+c^n}$

$\ge\dfrac{a^{m-n}+b^{m-n}+c^{m-n}}{3}\cdot\text{∑}\dfrac{a^n}{b^n+c^n}$

$\ge\text{∑}\dfrac{a^{m-n}+b^{m-n}+c^{m-n}}{2}$

We're done when $a=b=c$

P/s: This's discuss not A⁴ (Auto Ask Auto Answer)

Macedoi

24/07/2017 at 18:42

Votes Answers Follow

Given $a,b,c>0$. Prove that $\frac{a^3b}{3a+b}+\frac{b^3c}{3b+c}+\frac{c^3a}{3c+a}\ge \frac{a^2bc}{2a+b+c}+\frac{ab^2c}{a+2b+c}+\frac{abc^2}{a+b+2c}$

*)Source: https://olm.vn/hoi-dap/question/999979.html

I tried AM-GM but unsuccess

inequality

Ely Seon 05/08/2018 at 09:54

Considering the right expression we have $\dfrac{a^2bc}{2a+b+c}+\dfrac{ab^2c}{a+2b+c}+\dfrac{abc^2}{a+b+2c}$

$=abc\left(\dfrac{a}{a+b+a+c}+\dfrac{b}{a+b+b+c}+\dfrac{c}{a+c+b+c}\right)$

Apply Cauchy Schwarz inequality we have the formula $\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge\dfrac{1}{a+b}$

Apply this formula to the following expression we have

$\dfrac{a}{a+b+a+c}\le\dfrac{a}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)$$;\dfrac{b}{a+b+b+c}\le\dfrac{b}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)$

$\dfrac{c}{a+c+c+a}\le\dfrac{c}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{c+a}\right)$

$\Rightarrow\dfrac{a}{a+b+a+c}+\dfrac{b}{a+b+b+c}+\dfrac{c}{a+c+c+a}\le\dfrac{a+b}{4\left(a+b\right)}+\dfrac{b+c}{4\left(b+c\right)}+\dfrac{c+a}{4\left(c+a\right)}=\dfrac{3}{4}$

$\Rightarrow abc\left(\dfrac{a}{a+b+a+c}+\dfrac{b}{a+b+b+c}+\dfrac{c}{a+c+b+c}\right)\le\dfrac{3abc}{4}$

We now need to prove that $\dfrac{a^3b}{3a+b}+\dfrac{b^3c}{3b+c}+\dfrac{c^3a}{3c+a}\ge\dfrac{3abc}{4}$

$\Leftrightarrow\dfrac{a^3b}{3a+b}-\dfrac{abc}{4}+\dfrac{b^3c}{3b+c}-\dfrac{abc}{4}+\dfrac{c^3a}{3c+a}-\dfrac{abc}{4}\ge0$

$\Leftrightarrow\dfrac{a^3bc}{c\left(3a+b\right)}-\dfrac{abc}{4}+\dfrac{ab^3c}{a\left(3b+c\right)}-\dfrac{abc}{4}+\dfrac{c^3ab}{b\left(3c+a\right)}-\dfrac{abc}{4}\ge0$

$\Leftrightarrow abc\left[\dfrac{a^2}{c\left(3a+b\right)}-\dfrac{1}{4}\right]+abc\left[\dfrac{b^2}{a\left(3b+c\right)}-\dfrac{1}{4}\right]+abc\left[\dfrac{c^2}{b\left(3c+a\right)}-\dfrac{1}{4}\right]\ge0$

$\Leftrightarrow abc\left[\dfrac{a^2}{c\left(3a+b\right)}+\dfrac{b^2}{a\left(3b+c\right)}+\dfrac{c^2}{b\left(3c+a\right)}-\dfrac{3}{4}\right]\ge0$

$\Leftrightarrow\dfrac{a^2}{c\left(3a+b\right)}+\dfrac{b^2}{a\left(3b+c\right)}+\dfrac{c^2}{b\left(3c+a\right)}\ge\dfrac{3}{4}$

Apply Cauchy Schwarz inequality Engel form we have

$\dfrac{a^2}{c\left(3a+b\right)}+\dfrac{b^2}{a\left(3b+c\right)}+\dfrac{c^2}{b\left(3c+a\right)}\ge\dfrac{\left(a+b+c\right)^2}{4\left(ab+bc+ac\right)}$

According to the Cauchy's consequence, we have $\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)$

$\Rightarrow\dfrac{\left(a+b+c\right)^2}{4\left(ab+bc+ac\right)}\ge\dfrac{3\left(ab+bc+ac\right)}{4\left(ab+bc+ac\right)}=\dfrac{3}{4}$

$\Rightarrow\dfrac{a^2}{c\left(3a+b\right)}+\dfrac{b^2}{a\left(3b+c\right)}+\dfrac{c^2}{b\left(3c+a\right)}\ge\dfrac{3}{4}$

So now we have $\dfrac{a^3b}{3a+b}+\dfrac{b^3c}{3b+c}+\dfrac{c^3a}{3c+a}\ge\dfrac{3abc}{4}$

But $abc\left(\dfrac{a}{a+b+a+c}+\dfrac{b}{a+b+b+c}+\dfrac{c}{a+c+b+c}\right)\le\dfrac{3abc}{4}$

$\Rightarrow\dfrac{a^3b}{3a+b}+\dfrac{b^3c}{3b+c}+\dfrac{c^3a}{3c+a}\ge\dfrac{a^2bc}{2a+b+c}+\dfrac{ab^2c}{a+2b+c}+\dfrac{abc^2}{a+b+2c}$ ( things must be proven.)

Aim Egst

18/07/2017 at 18:21

Votes Answers Follow

For $a,x,y,z$ are positive real numbers satisfy $xyz=1$ and $a\ge 1$. Prove that $\dfrac{x^a}{y+z}+\dfrac{y^a}{x+z}+\dfrac{z^a}{x+y}\ge\dfrac{3}{2}$

-Like Nesbitt's inequality, i have a problem it's a general formula of Nesbitt's inequality, if necessary please inbox to me :)

inequality

Ace Legona

16/07/2017 at 12:57

Votes Answers Follow

Prove that with $n>3$ and $x_1;x_2;...;x_n>0$ satisfy $\Pi^n_{i=1}x_i=1$, so $\dfrac{1}{1+x_1+x_1x_2}+\dfrac{1}{1+x_2+x_2x_3}+...+\dfrac{1}{1+x_n+x_nx_1}>1$

- Source: The problem from Russia,2004

P/s: Spammer go away, Contraindication for young buffalo !!

inequality

FA KAKALOTS 28/01/2018 at 22:06

Yes but it's not mine, i need new method :)

Let x1=a2a1;x2=a3a2;...;xn=a1an

We need prove a1a1+a2+a3+a2a2+a3+a4+...+anan+a1+a2>1

It's right because n>3

and ai+ai+1+ai+2<a1+a2+...+an∀i
Aim Egst 18/07/2017 at 11:32

Yes but it's not mine, i need new method :)

Let $x_1=\dfrac{a_2}{a_1};x_2=\dfrac{a_3}{a_2};...;x_n=\dfrac{a_1}{a_n}$

We need prove $\dfrac{a_1}{a_1+a_2+a_3}+\dfrac{a_2}{a_2+a_3+a_4}+...+\dfrac{a_n}{a_n+a_1+a_2}>1$

It's right because $n>3$ and $a_i+a_{i+1}+a_{i+2}< a_1+a_2+...+a_n\forall i$
Summer Clouds moderators 18/07/2017 at 08:33

Ace Legona : you can share a answer of problem.

Kayasari Ryuunosuke Coodinator

16/07/2017 at 09:44

Votes Answers Follow

Give a,b,c are positive real number and a² + b² + c² = 3

Find Max_P know :

P = a + b + c - abc

inequality maxima-minima

Vũ Mạnh Hùng 05/12/2017 at 20:04

P=0 vì a+b+c=abc

suy ra abc -abc=0

answer :0
Vũ Mạnh Hùng 05/12/2017 at 19:59

P=0 vì a+b+c=abc

suy ra abc -abc=0

answer :0
Ace Legona 16/07/2017 at 12:46

wrong tag : maxima-minima

Let $a,b,c$ are positive real number satisfy $a+b+c=1$. Prove that $\dfrac{1}{\sqrt{\left (a^2+ab+b^2\right )\left (b^2+bc+c^2\right )}}+\dfrac{1}{\sqrt{\left (b^2+bc+c^2\right )\left (c^2+ca+a^2\right )}}+\dfrac{1}{\sqrt{\left (c^2+ca+a^2\right )\left (a^2+ab+b^2\right )}}\ge 4+\dfrac{8}{\sqrt{3}}$

inequality

Nguyen Huu Ai Linh 10/12/2017 at 07:08

That too long!

Nguyễn Nhật Minh

04/04/2017 at 18:29

Votes Answers Follow

Prove that:

B = $\dfrac{1}{2!}+\dfrac{5}{3!}+\dfrac{11}{4!}+...+\dfrac{n^2+n-1}{\left(n+1\right)!}< 2$.

Help me please! This is my homework and I must give it to my teacher tomorrow!

inequality Fraction

Nguyễn Nhật Minh

02/04/2017 at 11:25

Votes Answers Follow

Prove that E = $\dfrac{1}{3^3}+\dfrac{1}{4^3}+\dfrac{1}{5^3}+...+\dfrac{1}{n^3}< \dfrac{1}{12}$.

Fraction inequality

FA KAKALOTS 28/01/2018 at 22:08

For any natural number n > 1,we have :

(n - 1)n(n + 1) = n(n2 - 1) = n3 - n < n3

⇒1n3<1(n−1)n(n+1)

1(n−1)n(n+1)=1n.1(n−1)(n+1)

=1n.(n+1)−(n−1)(n−1)(n+1).12=12.1n.(1n−1−1n+1)

=12.(1(n−1)n−1n(n+1))

Now we have :

E < 12.3.4+13.4.5+14.5.6+...+1(n−1)n(n+1)

=12(12.3−13.4)+12(13.4−14.5)+12(14.5−15.6)+...+12(1(n−1)n−1n(n+1))

=12(12.3−1n(n+1))=112−12n(n+1)<112

Hence,E<112
Phan Thanh Tinh Coodinator 24/04/2017 at 13:50

For any natural number n > 1,we have :

(n - 1)n(n + 1) = n(n² - 1) = n³ - n < n³

$\Rightarrow\dfrac{1}{n^3}< \dfrac{1}{\left(n-1\right)n\left(n+1\right)}$

$\dfrac{1}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{n}.\dfrac{1}{\left(n-1\right)\left(n+1\right)}$

$=\dfrac{1}{n}.\dfrac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)\left(n+1\right)}.\dfrac{1}{2}=\dfrac{1}{2}.\dfrac{1}{n}.\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)$

$=\dfrac{1}{2}.\left(\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\right)$

Now we have :

E < $\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+\dfrac{1}{4.5.6}+...+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}$

$=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+\dfrac{1}{2}\left(\dfrac{1}{3.4}-\dfrac{1}{4.5}\right)+\dfrac{1}{2}\left(\dfrac{1}{4.5}-\dfrac{1}{5.6}\right)+...+\dfrac{1}{2}\left(\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\right)$

$=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{1}{12}-\dfrac{1}{2n\left(n+1\right)}< \dfrac{1}{12}$

Hence,$E< \dfrac{1}{12}$

inequality

Kayasari Ryuunosuke Coodinator

4778

Weekly ranking

Tags