Áp dụng BĐT Cosi cho 2 số dương ta có: \(x^2+yz\ge2\sqrt{x^2yz}=2x\sqrt{yz}\)
Tương tự: \(y^2+zx\ge2y\sqrt{zx},z^2+xy\ge2z\sqrt{xy}\)
Khi đó BĐT sẽ được chứng minh nếu ta chỉ ra được:
\(\frac{1}{2x\sqrt{yz}}+\frac{1}{2y\sqrt{zx}}+\frac{1}{2z\sqrt{xy}}\le\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)
\(\Leftrightarrow\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}{xyz}\le\frac{x+y+z}{xyz}\Leftrightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\le x+y+z\)
\(\Leftrightarrow\frac{1}{2}\left(\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{y}-\sqrt{z}\right)^2+\left(\sqrt{z}-\sqrt{x}\right)^2\right)\ge0\)(luôn đúng)
Vậy BĐT được chứng minh. Dấu "=" xảy ra khi \(x=y=z\)
Cách 2:
Ta chuẩn hóa xyz=1
BĐT viết lại là \(\frac{x}{x^3+1}+\frac{y}{y^3+1}+\frac{z}{z^3+1}\le\frac{1}{2}\left(x+y+z\right)\)
Ta sử dụng đánh giá
\(x-\frac{2x}{x^3+1}+\frac{3}{2}\ge\frac{9x^2}{2\left(x^2+x+1\right)}\)\(\Leftrightarrow\frac{\left(x-1\right)^2\left(2x^4+3x^2+7x+3\right)}{2\left(x^3+1\right)\left(x^2+x+1\right)}\ge0\)
Do vậy ta cần c/m \(\frac{x^2}{x^2+x+1}+\frac{y^2}{y^2+y+1}+\frac{z^2}{z^2+z+1}\ge1\)
ta có \(\left(x;y;z\right)\rightarrow\left(\frac{a^2}{bc};\frac{b^2}{ca};\frac{c^2}{ab}\right)\)
BĐT viết lại là \(\frac{a^4}{a^4+a^2bc+\left(bc\right)^2}+\frac{b^4}{b^4+b^2ca+\left(ca\right)^2}+\frac{c^4}{c^4+c^2ab+\left(ab\right)^2}\ge1\)
Theo bđt Cauchy-Schwarz ta có
\(VT\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^4+b^4+c^4+abc\left(a+b+c\right)+\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2}\)
Theo bđt AM-GM ta có
\(VT\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^4+b^4+c^4+2\left(ab\right)^2+2\left(bc\right)^2+2\left(ca\right)^2}=1\)
Dấu "=" xảy ra khi a=b=c=> x=y=z
Cách 5:
\(VP-VT=\frac{1}{4xyz}\Sigma\frac{\left(xy+xz-2yz\right)^2+yz\left(y-z\right)^2}{\left(y+z\right)\left(x^2+yz\right)}\ge0\)
Áp dụng BĐT Cosi cho 2 số dương x2 và yz, ta có:
\(x^2+yz\ge2\sqrt{x^2yz}=2x\sqrt{yz}\Rightarrow\frac{1}{x^2+yz}\le\frac{1}{2}\cdot\frac{1}{x\sqrt{yz}}\)
Tương tự ta có: \(\hept{\begin{cases}\frac{1}{y^2+xz}\le\frac{1}{2}\cdot\frac{1}{y\sqrt{xz}}\\\frac{1}{z^2+xy}\le\frac{1}{2}\cdot\frac{1}{z\sqrt{xy}}\end{cases}}\)
\(\Rightarrow\frac{1}{x^2+yz}+\frac{1}{y^2+xz}+\frac{1}{z^2+xy}\le\frac{1}{2}\left(\frac{1}{x\sqrt{zy}}+\frac{1}{y\sqrt{xz}}+\frac{1}{z\sqrt{xy}}\right)\left(1\right)\)
Ta có:\(\frac{1}{x\sqrt{yz}}+\frac{1}{y\sqrt{xz}}+\frac{1}{z\sqrt{xy}}=\frac{\sqrt{yz}+\sqrt{xz}+\sqrt{zx}}{xyz}\left(2\right)\)
Ta lại có: \(\sqrt{yz}+\sqrt{xz}+\sqrt{xy}\le x+y+z\left(3\right)\)
\(\Leftrightarrow2\sqrt{yz}+2\sqrt{xz}+2\sqrt{xy}\le2x+2y+2z\)
\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{z}-\sqrt{x}\right)^2+\left(\sqrt{y}-\sqrt{x}\right)^2\ge0\) (BĐT luôn đúng)
Dấu "=" xảy ra <=> x=y=z
Từ (2) (3) => \(\frac{1}{x\sqrt{yz}}+\frac{1}{y\sqrt{xz}}+\frac{1}{z\sqrt{xy}}\le\frac{x+y+z}{xyz}=\frac{1}{yz}+\frac{1}{xz}+\frac{1}{zx}\left(4\right)\)
Từ (1) (4) => \(\frac{1}{x^2+yz}+\frac{1}{y^2+xz}+\frac{1}{z^2+xy}\le\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)
