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\(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\)
Dấu "=" xảy ra khi \(x=y=z\)
Hoặc:
\(\frac{x^2}{y+z}+\frac{y+z}{4}\ge2\sqrt{\frac{x^2\left(y+z\right)}{4\left(y+z\right)}}=x\)
\(\frac{y^2}{x+z}+\frac{x+z}{4}\ge y\) ; \(\frac{z^2}{x+y}+\frac{x+y}{4}\ge z\)
Cộng vế với vế ta có đpcm
Sử dụng BĐT AM-GM, ta có:
\(x^3+y^2\ge2yx\sqrt{x}\)
\(\Rightarrow\frac{2\sqrt{x}}{x^3+y^2}\le\frac{2\sqrt{x}}{2yx\sqrt{x}}=\frac{1}{xy}\)
Tương tự cộng lại suy ra:
\(VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
\(\Leftrightarrow\frac{x^2}{y+z}-\frac{z^2}{y+z}+\frac{z^2}{x+y}-\frac{y^2}{x+y}+\frac{y^2}{x+z}-\frac{x^2}{x+z}\ge0\)
\(\Leftrightarrow\left(\frac{x^2}{y+z}-\frac{x^2}{x+z}\right)+\left(\frac{y^2}{x+z}-\frac{y^2}{x+y}\right)+\left(\frac{z^2}{x+y}-\frac{z^2}{y+z}\right)\ge0\)
\(\Leftrightarrow x^2\left(\frac{1}{y+z}-\frac{1}{x+z}\right)+y^2\left(\frac{1}{x+z}-\frac{1}{x+y}\right)+z^2\left(\frac{1}{x+y}-\frac{1}{y+z}\right)\ge0\)
\(\Leftrightarrow x^2\left(\frac{x-y}{\left(y+z\right)\left(x+z\right)}\right)+y^2\left(\frac{y-z}{\left(x+z\right)\left(x+y\right)}\right)+z^2\left(\frac{z-x}{\left(x+y\right)\left(y+z\right)}\right)\ge0\)
\(\Leftrightarrow x^2\left(x-y\right)\left(x+y\right)+y^2\left(y-z\right)\left(y+z\right)+z^2\left(z-x\right)\left(z+x\right)\ge0\)
\(\Leftrightarrow x^2\left(x^2-y^2\right)+y^2\left(y^2-z^2\right)+z^2\left(z^2-x^2\right)\ge0\)
\(x^4-x^2y^2+y^4-y^2z^2+z^4-z^2x^2\ge0\)
\(\Leftrightarrow2x^4-2x^2y^2+2y^4-2y^2z^2+2z^4-2z^2x^2\ge0\)
\(\Leftrightarrow\left(x^4-2x^2y^2+y^4\right)+\left(y^4-2y^2z^2+z^4\right)+\left(z^4-2z^2x^2+x^4\right)\ge0\)
\(\Leftrightarrow\left(x^2-y^2\right)^2+\left(y^2-z^2\right)^2+\left(z^2-x^2\right)^2\ge0\)(đúng)
\(VT=x^2+y^2+z^2+3-\frac{y^2\left(x^2+1\right)}{y^2+1}-\frac{z^2\left(y^2+1\right)}{z^2+1}-\frac{x^2\left(z^2+1\right)}{x^2+1}\)
\(\le x^2+y^2+z^2+3-\frac{y^2\left(x^2+1\right)+z^2\left(y^2+1\right)+x^2\left(z^2+1\right)}{2}\)
\(\le\frac{x^2+y^2+z^2}{2}+3-\frac{x^2y^2+y^2z^2+z^2x^2}{2}\)
\(\le\frac{x^2+y^2+z^2}{2}+3\)
Mặt khác ta có: \(x^2+y^2+z^2=1-2\left(xy+yz+zx\right)\le1\)
\(\Rightarrow VT\le\frac{7}{2}\).Dấu "=" xảy ra tại \(\left(0;0;1\right)\) và các hoán vị của nó
Với \(\hept{\begin{cases}x,y,z\ge0\\x+y+z=1\end{cases}}\), ta cần chứng minh: \(\frac{x^2+1}{y^2+1}+\frac{y^2+1}{z^2+1}+\frac{z^2+1}{x^2+1}\le\frac{7}{2}\)
\(\Leftrightarrow2\Sigma_{cyc}\left(x^2+1\right)^2\left(z^2+1\right)\le7\left(x^2+1\right)\left(y^2+1\right)\left(z^2+1\right)\) \(\Leftrightarrow2\Sigma_{cyc}\left(x^4z^2+x^4+2x^2z^2+2x^2+z^2+1\right)\)\(\le7\left(x^2y^2z^2+x^2+y^2+z^2+x^2y^2+y^2z^2+z^2x^2+1\right)\)
\(\Leftrightarrow2\left(x^4+y^4+z^4\right)+2\left(x^4z^2+y^4x^2+z^4y^2\right)\)\(\le7x^2y^2z^2+3\left(x^2y^2+y^2z^2+z^2x^2\right)+x^2+y^2+z^2+1\)
\(\Leftrightarrow\left[x^2+y^2+z^2+x+y+z-2\left(x^4+y^4+z^4\right)\right]\)\(+7x^2y^2z^2+3\left(x^2y^2+y^2z^2+z^2x^2\right)-2\left(x^4z^2+y^4x^2+z^4y^2\right)\ge0\)
\(\Leftrightarrow\text{}\Sigma_{cyc}x^2\left(1-x^2\right)+\Sigma_{cyc}x\left(1-x^3\right)+7x^2y^2z^2\)\(+\left(x^2z^2+y^2x^2+z^2y^2\right)+2\Sigma x^2z^2\left(1-x^2\right)\ge0\)
(Đúng do \(x,y,z\in\left[0;1\right]\))
Đẳng thức xảy ra khi \(\left(x,y,z\right)=\left(1;0;0\right)\)và các hoán vị