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Ta có: \(\dfrac{x}{x^2+1+y^2+1}\le\dfrac{x}{2\sqrt{\left(x^2+1\right)\left(y^2+1\right)}}\le\dfrac{1}{4}\left(\dfrac{x^2}{x^2+1}+\dfrac{1}{y^2+1}\right)\)
Tương tự: \(\dfrac{y}{y^2+z^2+2}\le\dfrac{1}{4}\left(\dfrac{y^2}{y^2+1}+\dfrac{1}{z^2+1}\right)\) ; \(\dfrac{z}{z^2+x^2+2}\le\dfrac{1}{4}\left(\dfrac{z^2}{z^2+1}+\dfrac{1}{x^2+1}\right)\)
Cộng vế với vế:
\(P\le\dfrac{1}{4}\left(\dfrac{x^2}{x^2+1}+\dfrac{1}{x^2+1}+\dfrac{y^2}{y^2+1}+\dfrac{1}{y^2+1}+\dfrac{z^2}{z^2+1}+\dfrac{1}{z^2+1}\right)=\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Áp dụng BĐT cosi:
\(\dfrac{x^2}{y+z}+\dfrac{y+z}{4}\ge2\sqrt{\dfrac{x^2\left(y+z\right)}{4\left(y+z\right)}}=\dfrac{2x}{2}=x\)
Cmtt \(\dfrac{y^2}{x+z}+\dfrac{x+z}{4}\ge y;\dfrac{z^2}{x+y}+\dfrac{x+y}{4}\ge z\)
Cộng VTV 3 BĐT trên:
\(\Leftrightarrow\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}+\dfrac{2\left(x+y+z\right)}{4}\ge x+y+z\\ \Leftrightarrow\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\ge x+y+z-\dfrac{x+y+z}{2}=\dfrac{x+y+z}{2}\)
Dấu \("="\Leftrightarrow x=y=z\)
\(VT=\dfrac{\left(\dfrac{1}{z}\right)^2}{\dfrac{1}{x}+\dfrac{1}{y}}+\dfrac{\left(\dfrac{1}{x}\right)^2}{\dfrac{1}{y}+\dfrac{1}{z}}+\dfrac{\left(\dfrac{1}{y}\right)^2}{\dfrac{1}{x}+\dfrac{1}{z}}\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}{2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Dâu "=" xảy ra khi \(x=y=z\)
Lời giải:
\(A=\left(\frac{x}{y-z}+\frac{y}{z-x}+\frac{z}{x-y}\right)\left(\frac{1}{y-z}+\frac{1}{z-x}+\frac{1}{x-y}\right)-\frac{x}{(y-z)(z-x)}-\frac{x}{(y-z)(x-y)}-\frac{y}{(z-x)(x-y)}-\frac{y}{(z-x)(y-z)}-\frac{z}{(x-y)(y-z)}-\frac{z}{(x-y)(z-x)}\)
\(=0-\frac{x(x-y)+x(z-x)+y(y-z)+y(x-y)+z(z-x)+z(y-z)}{(x-y)(y-z)(z-x)}\)
\(=0-\frac{x^2+xz+y^2+xy+z^2+zy-(xy+x^2+yz+y^2+zx+z^2)}{(x-y)(y-z)(z-x)}=0-\frac{0}{(x-y)(y-z)(z-x)}=0\)
\(\sum\dfrac{1}{x}\cdot\sum\dfrac{x}{y^2}\ge\sum^2\dfrac{1}{x}\)(bunhia)
\(x+y+z=xyz\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
\(VT\le\dfrac{x}{2\sqrt{x^2yz}}+\dfrac{y}{2\sqrt{y^2zx}}+\dfrac{z}{2\sqrt{z^2xy}}\)
\(VT\le\dfrac{1}{2}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\le\dfrac{1}{2}\sqrt{3\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)}=\dfrac{\sqrt{3}}{2}\)
Dấu "=" xảy ra khi \(x=y=z=\sqrt{3}\)
\(x+y+z=xyz\Leftrightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
Đặt \(\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)=\left(a;b;c\right)\Rightarrow ab+bc+ca=1\)
\(Q=\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}+2\left(a^2+b^2+c^2\right)\)
\(Q\ge\dfrac{\left(a+b+c\right)^2}{a+b+c}+2\left(ab+bc+ca\right)=a+b+c+2\)
\(Q\ge\sqrt{3\left(ab+bc+ca\right)}+2=2+\sqrt{3}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\) hay \(x=y=z=\sqrt{3}\)