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:

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

maxima-minima

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