Ta có:

\(P=\left(\frac{1}{x^2+y^2+z^2}+\frac{1}{3xy}+\frac{1}{3yz}+\frac{1}{3zx}\right)+\frac{5}{12}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)

\(\ge\frac{\left(1+1+1+1\right)^2}{x^2+y^2+z^2+3xy+3yz+3zx}+\frac{5}{12}.\frac{\left(1+1+1\right)^2}{xy+yz+zx}\)

\(=\frac{16}{\left(x+y+z\right)^2+\left(xy+yz+zx\right)}+\frac{5}{12}.\frac{9}{xy+yz+zx}\)

\(\ge\frac{16}{\left(x+y+z\right)^2+\frac{\left(x+y+z\right)^2}{3}}+\frac{5}{12}.\frac{9}{\frac{\left(x+y+z\right)^2}{3}}\)

\(=\frac{93}{4\left(x+y+z\right)^2}=\frac{93}{4\left(2019\right)^2}\)

Dấu "=" xảy ra <=> x = y = z = 2019/3.

Ta co A = 2(x+y)+\(\frac{2}{x+y}\)\(\ge2\sqrt{2\left(x+y\right).\frac{2}{x+y}}\)^{=4          khi x=y =\(\frac{1}{2}\)}

\(\frac{x^2}{y+1}+\frac{y+1}{4}\ge x;\frac{y^2}{z+1}+\frac{z+1}{4}\ge y;\frac{z^2}{x+1}+\frac{x+1}{4}\ge z\)

\(\Rightarrow VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{4}.2=\frac{3}{2}\)

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\(Q=\frac{x}{1+y^2}+\frac{y}{1+z^2}+\frac{z}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+z^2}+\frac{1}{1+x^2}\)

Ta có \(\frac{x}{1+y^2}=\frac{x\left(1+y^2\right)-xy^2}{1+y^2}=x-\frac{xy^2}{1+y^2}\ge x-\frac{xy^2}{2y}=x-\frac{xy}{2}\)

Tương tự \(\frac{y}{1+z^2}\ge y-\frac{yz}{2}\)

                    \(\frac{z}{1+x^2}\ge z-\frac{zx}{2}\)

Lại có \(\frac{1}{1+y^2}=\frac{y^2+1-y^2}{1+y^2}=1-\frac{y^2}{1+y^2}\ge1-\frac{y^2}{2y}=1-\frac{y}{2}\)

Tương tự \(\frac{1}{1+x^2}\ge1-\frac{x}{2}\)

\(\frac{1}{1+z^2}\ge1-\frac{z}{2}\)

Cộng từng vế các bđt trên ta được 

\(Q\ge\left(x+y+z\right)-\frac{xy+yz+zx}{2}+3-\frac{x+y+z}{2}\)\(=\frac{9}{2}-\frac{3}{2}=3\)

Dấu "=" xảy ra khi x=y=z=1

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# Hok tốt !

\(x+y=1\Rightarrow\hept{\begin{cases}1-x=y\\1-y=x\end{cases}}\)

thay vào A ta được : \(A=\frac{1-y}{\sqrt{y}}+\frac{1-x}{\sqrt{x}}\)

\(\Rightarrow A=\frac{1}{\sqrt{y}}-\sqrt{y}+\frac{1}{\sqrt{x}}-\sqrt{x}\)

\(\Rightarrow A=\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)-\left(\sqrt{x}+\sqrt{y}\right)\)

áp dụng \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ta có : \(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\ge\frac{4}{\sqrt{x}+\sqrt{y}}\)

áp dụng \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\) ta có : \(\left(\sqrt{x}+\sqrt{y}\right)^2\le2\left(\sqrt{x}^2+\sqrt{y}^2\right)=2\)

\(\Rightarrow\sqrt{x}+\sqrt{y}\le\sqrt{2}\)

\(\Rightarrow A\ge\sqrt{8}-\sqrt{2}=\sqrt{2}\)

dấu = xảy ra khi a=y=1/2