maxima-minima
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)
-
FA KAKALOTS 28/01/2018 at 22:06i have a solution but it's ugly
If a=b=1
and c=0 then we get a value 2+1√2
We'll prove that it's a minimal value. Thus, we need to prove that
√ab+bc+caa2+b2+√ab+bc+cab2+c2+√ab+bc+caa2+c2≥2+1√2
WLOG c=min{a,b,c}
. Hence:
ab+ac+bca2+b2−(a+c)(b+c)(a+c)2+(b+c)2=c(a+b+2c)(2ab+ac+bc)a2+b2)((a+c)2+(b+c)2≥0
Similarab+ac+bca2+c2−b+ca+c=c(2ab+ac−c2)(a+c)(a2+c2)≥0
And ab+ac+bcb2+c2−a+cb+c=c(2ab+bc−c2)(b+c)(b2+c2)≥0
Let a+cb+c=x2;b+ca+c=y2(x,y>0)
⇒xy=1
and we have:
x+y+1√x2+y2≥2+1√2
⇔x+y−2√xy≥1√2−1√x2+y2
⇔(√x−√y)2≥(x−y)2√2(x2+y2)(√x2+y2+√2)
⇔√2(x2+y2)(√x2+y2+√2)≥(√x+√y)2
By Cauchy-Schwarz's ine we have:
√2(x2+y2)=√(12+12)(x2+y2)≥x+y
=12(12+12)((√x)2+(√y)2)≥12(√x+√y)2
Thus, it's enough to prove that √x2+y2+√2≥2
It's true by AM-GM √x2+y2+√2≥√2xy+√2=2√2>2
-
AL 08/08/2017 at 13:15i 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 20:10tôi để lại số quả táo là:
Kayasari Ryuunosuke Coodinator
16/07/2017 at 09:44-
-
-
Ace Legona 16/07/2017 at 12:46
wrong tag : maxima-minima
