Question

Cho x,y,z >0 tm 1/xy +1/yz + 1/xz =1

Tính Max Q = \(\frac{x}{\sqrt{yz\left(1+x^2\right)}}+\frac{y}{\sqrt{xz\left(1+y^2\right)}}+\frac{z}{\sqrt{xy\left(1+z^2\right)}}\)

Chị @Akai Haruma giúp e làm bài này đc k ạ!!!

Answer 1

Lời giải:

Đặt \(\left(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\right)=(a,b,c)\). Bài toán đã cho trở thành:

Cho $a,b,c>0$ thỏa mãn \(ab+bc+ac=1\)

Tính max của \(Q=\frac{\sqrt{bc}}{\sqrt{a^2+1}}+\frac{\sqrt{ac}}{\sqrt{b^2+1}}+\frac{\sqrt{ab}}{\sqrt{c^2+1}}\)

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Vì $ab+bc+ac=1$ nên:

\(Q=\sqrt{\frac{bc}{a^2+ab+bc+ac}}+\sqrt{\frac{ac}{b^2+ab+bc+ac}}+\sqrt{\frac{ab}{c^2+ab+bc+ac}}\)

\(=\sqrt{\frac{bc}{(a+b)(a+c)}}+\sqrt{\frac{ac}{(b+c)(b+a)}}+\sqrt{\frac{ab}{(c+a)(c+b)}}\)

Áp dụng BĐT Cauchy:

\(Q\leq \frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)+\frac{1}{2}\left(\frac{a}{b+a}+\frac{c}{b+c}\right)+\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{b+c}\right)\)

\(Q\leq \frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\)

Vậy \(Q_{\max}=\frac{3}{2}\)

Answer 2

\(x,y,z>0:\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\left(1\right)\)

\(Q=\frac{x}{\sqrt{yz\left(1+x^2\right)}}+\frac{y}{\sqrt{xz\left(1+y^2\right)}}+\frac{z}{\sqrt{xy\left(1+z^2\right)}}\)

Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\left(a,b,c>0\right)\)

\(Q=\sqrt{\frac{\frac{1}{yz}}{1+\frac{1}{x^2}}}+\sqrt{\frac{\frac{1}{xz}}{1+\frac{1}{y^2}}}+\sqrt{\frac{\frac{1}{xy}}{1+\frac{1}{z^2}}}\\ =\sqrt{\frac{bc}{1+a^2}}+\sqrt{\frac{ac}{1+b^2}}+\sqrt{\frac{ab}{1+c^2}}\)

\(\left(1\right)\Leftrightarrow ab+bc+ca=1\\ \Rightarrow a^2+1=a^2+ab+ac+bc=\left(a+b\right)\left(a+c\right)\\ \Rightarrow\sqrt{\frac{bc}{1+a^2}}=\sqrt{\frac{b}{a+b}}.\sqrt{\frac{c}{a+c}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\)

Tương tự: \(\sqrt{\frac{ca}{1+b^2}}\le\frac{1}{2}\left(\frac{c}{b+c}+\frac{a}{b+a}\right)\)

\(\sqrt{\frac{ab}{1+c^2}}\le\frac{1}{2}\left(\frac{a}{a+c}+\frac{b}{b+c}\right)\\ \Rightarrow Q\le\frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\)

(Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\Leftrightarrow x=y=z=\sqrt{3}\))