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Ta có: \(x^2\left(y+z\right)\ge x^2.2\sqrt{yz}=2\sqrt{x^4}.\sqrt{\frac{1}{x}}=2x\sqrt{x}\)(Áp dụng BĐT Cô - si cho 2 số dương y,z và sử dụng giả thiết xyz = 1)
Hoàn toàn tương tự: \(y^2\left(z+x\right)\ge2y\sqrt{y};z^2\left(x+y\right)\ge2z\sqrt{z}\)
Do đó \(P=\frac{x^2\left(y+z\right)}{y\sqrt{y}+2z\sqrt{z}}+\frac{y^2\left(z+x\right)}{z\sqrt{z}+2x\sqrt{x}}+\frac{z^2\left(x+y\right)}{x\sqrt{x}+2y\sqrt{y}}\)
\(\ge\frac{2x\sqrt{x}}{y\sqrt{y}+2z\sqrt{z}}+\frac{2y\sqrt{y}}{z\sqrt{z}+2x\sqrt{x}}+\frac{2z\sqrt{z}}{x\sqrt{x}+2y\sqrt{y}}\)
Đặt \(a=x\sqrt{x}+2y\sqrt{y}\), \(b=y\sqrt{y}+2z\sqrt{z}\), \(c=z\sqrt{z}+2x\sqrt{x}\)
Suy ra: \(x\sqrt{x}=\frac{4c+a-2b}{9}\), \(y\sqrt{y}=\frac{4a+b-2c}{9}\), \(z\sqrt{z}=\frac{4b+c-2a}{9}\)
Do đó \(P\ge\frac{2}{9}\left(\frac{4c+a-2b}{b}+\frac{4a+b-2c}{c}+\frac{4b+c-2a}{a}\right)\)
\(=\frac{2}{9}\left[4\left(\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\right)+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-6\right]\)
\(\ge\frac{2}{9}\left[4.3\sqrt[3]{\frac{c}{b}.\frac{a}{c}.\frac{b}{a}}+3\sqrt[3]{\frac{a}{b}.\frac{b}{c}.\frac{c}{a}}-6\right]\)(Áp dụng BĐT Cô - si cho 3 số dương)
\(=\frac{2}{9}\left[4.3+3-6\right]=2\)
Vậy \(P\ge2\)
Đẳng thức xảy ra khi x = y = z = 1
theo bat dang thuc C-S ta co
\(P\le\frac{x}{x+\sqrt{xy}+\sqrt{xz}}+\frac{y}{y+\sqrt{yz}+\sqrt{yx}}+\frac{z}{z+\sqrt{zx}+\sqrt{zy}}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\frac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)
Vay GTLN cua P la 1 dau = khi x=y=z
\(\frac{x}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{z}\right)}+\frac{y}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{x}\right)}+\)\(\frac{z}{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{z}-\sqrt{y}\right)}\)
\(=-\frac{x}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{z}-\sqrt{x}\right)}-\frac{y}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)\(-\frac{z}{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)
\(=\frac{-x\left(\sqrt{y}-\sqrt{z}\right)-y\left(\sqrt{z}-\sqrt{x}\right)-z\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
\(=\frac{-x\sqrt{y}+x\sqrt{z}-y\sqrt{z}+y\sqrt{x}-z\sqrt{x}+z\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
\(=\frac{-\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)+\sqrt{z}\left(x-y\right)-z\left(\sqrt{x}-y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
\(=\frac{-\sqrt{xy}+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)-z}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
\(=\frac{-\sqrt{xy}+\sqrt{xz}+\sqrt{yz}-z}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
\(=\frac{\sqrt{y}\left(\sqrt{z}-\sqrt{x}\right)-\sqrt{z}\left(\sqrt{z}-\sqrt{x}\right)}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
\(=\frac{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{y}-\sqrt{z}\right)}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
\(A=\frac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}+\frac{\left(x+z\right)\sqrt{\left(x+y\right)\left(y+z\right)}}{y}+\frac{\left(x+y\right)\sqrt{\left(y+z\right)\left(x+z\right)}}{z}.\)
Áp dụng bất đẳng thức Bunhiacopski ta có
\(\left(x+y\right)\left(x+z\right)\ge\left(x+\sqrt{yz}\right)^2\)
Tương tự \(\left(x+y\right)\left(y+z\right)\ge\left(y+\sqrt{xz}\right)^2\)
\(\left(y+z\right)\left(x+z\right)\ge\left(z+\sqrt{xy}\right)^2\)
\(\Rightarrow A\ge\frac{\left(y+z\right)\left(x+\sqrt{yz}\right)}{x}+\frac{\left(x+z\right)\left(y+\sqrt{xz}\right)}{y}+\frac{\left(x+y\right)\left(z+\sqrt{xy}\right)}{z}\)
hay \(A\ge2\left(x+y+z\right)+\frac{\sqrt{yz}\left(y+z\right)}{x}+\frac{\left(x+z\right)\sqrt{xz}}{y}+\frac{\left(x+y\right)\sqrt{xy}}{z}\)
\(\Leftrightarrow A\ge2\left(x+y+z\right)+\frac{yz\sqrt{yz}\left(y+z\right)}{xyz}+\frac{xz\sqrt{xz}\left(x+z\right)}{xyz}+\frac{xy\sqrt{xy}\left(x+y\right)}{xyz}\)
Đặt \(M=\frac{yz\sqrt{yz}\left(y+z\right)}{xyz}+\frac{xz\sqrt{xz}\left(x+z\right)}{xyz}+\frac{xy\sqrt{xy}\left(x+y\right)}{xyz}\)
Ta có \(\left(x,y,z\right)\rightarrow\left(a^2,b^2,c^2\right)\)
Khi đó \(M=\frac{a^3b^3\left(a^2+b^2\right)+b^3c^3\left(b^2+c^2\right)+c^3a^3\left(a^2+c^2\right)}{a^2b^2c^2}\)
ÁP DỤNG BĐT AM-GM ta có
\(a^5b^3+a^3b^5\ge2\sqrt{a^8b^8}=2a^4b^4\)
\(b^5c^3+b^3c^5\ge2\sqrt{b^8c^8}=2b^4c^4\)
\(a^5c^3+a^3c^5\ge2\sqrt{a^8c^8}=2a^4c^4\)
Cộng từng vế ta được
\(a^3b^3\left(a^2+b^2\right)+b^3c^3\left(b^2+c^2\right)+c^3a^3\left(a^2+c^2\right)\ge2\left(a^4b^4+b^4c^4+c^4a^4\right)\)
\(\ge2a^2b^2c^2\left(a^2+b^2+c^2\right)\)
\(\Rightarrow M\ge2\left(a^2+b^2+c^2\right)=2\left(x+y+z\right)\)
\(\Rightarrow A\ge4\left(x+y+z\right)=4\sqrt{2019}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{\sqrt{2019}}{3}\)
Đặt \(\sqrt{x}=a\) , \(\sqrt{y}=b\) , \(\sqrt{z}=c\)
Suy ra \(P=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(b-a\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(=-\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Xét tử : \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)=a^2\left[-\left(a-b\right)-\left(c-a\right)\right]+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(c^2-a^2\right)+\left(c-a\right)\left(b^2-a^2\right)=\left(a-b\right)\left(c-a\right)\left(c+a\right)+\left(c-a\right)\left(b-a\right)\left(b+a\right)\)
\(=\left(a-b\right)\left(c-a\right)\left(c+a-a-b\right)=\left(a-b\right)\left(c-a\right)\left(c-b\right)\)
Suy ra \(P=-\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)