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\(\frac{6}{2xy+2yz+2zx}+\frac{2}{x^2+y^2+z^2}\ge\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=8+4\sqrt{3}>14\)
Dấu "=" không xảy ra
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 \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}=6\)
\(yz\le\frac{1}{2}\left(y^2+z^2\right)=\frac{1}{2}\left(3-x^2\right)\)
\(\Rightarrow3-yz\ge3-\frac{1}{2}\left(3-x^2\right)=\frac{3}{2}+\frac{1}{2}x^2\)
\(\Rightarrow\frac{x}{3-yz}\le\frac{x}{\frac{3}{2}+\frac{1}{2}x^2}=\frac{2x}{x^2+3}\)
Làm tương tự và cộng lại ta có: \(VT\le2\left(\frac{x}{x^2+3}+\frac{y}{y^2+3}+\frac{z}{z^2+3}\right)\)
Ta sẽ chứng minh: với mọi \(0< x^2< 3\) ta luôn có: \(\frac{x}{x^2+3}\le\frac{x^2+3}{16}\)
Thật vậy, BĐT tương đương:
\(16x\le\left(x^2+3\right)^2\Leftrightarrow\left(x-1\right)^2\left(x^2+2x+9\right)\ge0\) (luôn đúng)
Tương tự: \(\frac{y}{y^2+3}\le\frac{y^2+3}{16}\) ; \(\frac{z}{z^2+3}\le\frac{z^2+3}{16}\)
Cộng vế với vế:
\(VT\le2.\frac{x^2+y^2+z^2+9}{16}=\frac{3}{2}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
\(\sqrt{\frac{xy}{xy+z}}=\sqrt{\frac{xy}{xy+z\left(x+y+z\right)}}=\sqrt{\frac{xy}{\left(x+z\right)\left(y+z\right)}}\le\frac{1}{2}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)\)
Tương tự: \(\sqrt{\frac{yz}{yz+x}}\le\frac{1}{2}\left(\frac{y}{x+y}+\frac{z}{x+z}\right)\) ; \(\sqrt{\frac{zx}{zx+y}}\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{z}{y+z}\right)\)
Cộng vế với vế ta có đpcm
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
\(VT=\sum\frac{x}{3-yz}\le\sum\frac{2x}{6-\left(y^2+z^2\right)}=\sum\frac{2x}{x^2+x^2+y^2+z^2}\le\sum\frac{x^2+1}{x^2+1+2}\)
\(VT\le\frac{1}{4}\sum\left(\frac{x^2+1}{x^2+1}+\frac{x^2+1}{2}\right)=\frac{1}{4}\left(3+\frac{x^2+y^2+z^2+3}{2}\right)=\frac{3}{2}\)