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Áp dụng bất đẳng thức Cauchy :
\(\frac{x^4}{y^2\left(x+z\right)}+\frac{y^2}{2x}+\frac{x+z}{4}\ge3\sqrt[3]{\frac{x^4\cdot y^2\cdot\left(x+z\right)}{y^2\cdot\left(x+z\right)\cdot2x\cdot4}}=3\sqrt[3]{\frac{x^3}{8}}=\frac{3x}{2}\)
Tương tự ta cũng có :
\(\frac{y^4}{z^2\left(x+y\right)}+\frac{z^2}{2y}+\frac{x+y}{4}\ge\frac{3y}{2}\)
\(\frac{z^4}{x^2\left(y+z\right)}+\frac{x^2}{2z}+\frac{y+z}{4}\ge\frac{3z}{2}\)
Cộng theo vế ta được :
\(VT+\left(\frac{y^2}{2x}+\frac{z^2}{2y}+\frac{x^2}{2z}\right)+\frac{2\left(x+y+z\right)}{4}\ge\frac{3x}{2}+\frac{3y}{2}+\frac{3z}{2}\)
\(\Leftrightarrow VT+\frac{1}{2}\left(\frac{y^2}{x}+\frac{z^2}{y}+\frac{x^2}{z}\right)+\frac{1}{2}\left(x+y+z\right)\ge\frac{3}{2}\left(x+y+z\right)\)
\(\Leftrightarrow VT+\frac{1}{2}\cdot\frac{\left(x+y+z\right)^2}{x+y+z}+\frac{1}{2}\left(x+y+z\right)\ge\frac{3}{2}\left(x+y+z\right)\)
\(\Leftrightarrow VT+\frac{1}{2}\left(x+y+z\right)+\frac{1}{2}\left(x+y+z\right)\ge\frac{3}{2}\left(x+y+z\right)\)
\(\Leftrightarrow VT\ge\frac{x+y+z}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)
Áp dụng BĐT Cô - si cho 3 bộ số không âm
\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(xz+1\right)^2}{x^2y^2z^2\left(yz+1\right)\left(xz+1\right)\left(xy+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
Xét \(3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
\(=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{xz+1}{z}\right)}\)
\(=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
Áp dụng BĐT Cô - si
\(\Rightarrow\left\{\begin{matrix}y+\frac{1}{x}\ge2\sqrt{\frac{y}{x}}\\z+\frac{1}{y}\ge2\sqrt{\frac{z}{y}}\\x+\frac{1}{z}\ge2\sqrt{\frac{x}{z}}\end{matrix}\right.\)
\(\Rightarrow\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)\ge8\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge3\sqrt[3]{8}\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge6\)
\(\Leftrightarrow3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\ge6\)
Mà \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge6\)
Vậy GTNN của \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}=6\)
Lời giải:
Đặt $\frac{x+y}{x-y}=a; \frac{y+z}{y-z}=b; \frac{z+x}{z-x}=c$
Bằng phép biến đổi tương đương cơ bản, ta chỉ ra được:
$ab+bc+ac=-1$
$\Leftrightarrow (a+b+c)^2-(a^2+b^2+c^2)=-2$
$\Leftrightarrow a^2+b^2+c^2=(a+b+c)^2+2\geq 2$
Ta sẽ đi chứng minh $a^{2020}+b^{2020}+c^{2020}>\frac{2^{1010}{3^{1009}}$
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Áp dụng BĐT AM-GM cho các số không âm:
\(\frac{a^{2020}}{a^{2020}+b^{2020}+c^{2020}}+\frac{1}{3}+\frac{1}{3}+....+\frac{1}{3}\geq 1010\sqrt[1010]{\frac{a^{2020}}{(a^{2020}+b^{2020}+c^{2020}).3^{1009}}}\)
\(\frac{b^{2020}}{a^{2020}+b^{2020}+c^{2020}}+\frac{1}{3}+\frac{1}{3}+....+\frac{1}{3}\geq 1010\sqrt[1010]{\frac{b^{2020}}{(a^{2020}+b^{2020}+c^{2020}).3^{1009}}}\)
\(\frac{c^{2020}}{a^{2020}+b^{2020}+c^{2020}}+\frac{1}{3}+\frac{1}{3}+....+\frac{1}{3}\geq 1010\sqrt[1010]{\frac{c^{2020}}{(a^{2020}+b^{2020}+c^{2020}).3^{1009}}}\)
Cộng theo vế và thu gọn: $a^2+b^2+c^2\leq \sqrt[1010]{(a^{2020}+b^{2020}+c^{2020}).3^{1009}}$
$\Rightarrow a^{2020}+b^{2020}+c^{2020}\geq \frac{(a^2+b^2+c^2)^{1010}}{3^{1009}}\geq \frac{2^{1010}}{3^{1009}}$ do $a^2+b^2+c^2\geq 2$
Dấu "=" xảy ra khi $a=b=c$ và $a^2+b^2+c^2=2$. Điều này không được vì $x,y,z$ đôi một khác nhau làm $a,b,c$ đôi một khác nhau
Ta có đpcm.
Akai Haruma dạ giúp em bài này vs ạ !!!