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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 \("="\Leftrightarrow x=y=z\)
Đặt \(\dfrac{x-y}{z}=m,\dfrac{y-z}{x}=n,\dfrac{z-x}{y}=p\), ta có:
\(\left(m+n+p\right)\left(\dfrac{1}{m}+\dfrac{1}{n}+\dfrac{1}{p}\right)=3+\dfrac{n+p}{m}+\dfrac{p+m}{n}+\dfrac{m+n}{p}\)
Tính \(\dfrac{n+p}{m}\) theo x, y, z ta được:
\(\dfrac{n+p}{m}=\dfrac{z}{x-y}.\dfrac{y^2-yz+xz-x^2}{xy}=\dfrac{z}{xy}\left(-x-y+x\right)\)
\(=\dfrac{z}{xy}\left(-x-y-z+2z\right)=\dfrac{2x^2}{xy}\) vì \(\left(x+y+z\right)=0\)
Tương tự: \(\dfrac{m+p}{n}=\dfrac{2x^2}{yz}.\dfrac{m+n}{p}=\dfrac{2y^2}{xz}\)
Vậy \(\left(m+n+p\right)\left(\dfrac{1}{m}+\dfrac{1}{n}+\dfrac{1}{p}\right)=3+\dfrac{2\left(x^3+y^3+z^3\right)}{xyz}=3+\dfrac{2.3xyz}{xyz}=3+6=9\)
Cho x, y, z > 0 và x+y+z=1.
CMR : \(\dfrac{1-x^2}{x+yz}+\dfrac{1-y^2}{y+zx}+\dfrac{1-z^2}{z+xy}\ge6\)
\(VT=\left(x^2+y^2+z^2\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)=3+\dfrac{x^2+y^2}{z^2}+z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\)
\(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}>=2\cdot\sqrt{\dfrac{y^2}{x^2}\cdot\dfrac{x^2}{y^2}}=2\)
=>\(VT>=5+\left(\dfrac{x^2}{z^2}+\dfrac{z^2}{16x^2}\right)+\left(\dfrac{y^2}{z^2}+\dfrac{z^2}{16y^2}\right)+\dfrac{15}{16}z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\)
\(\dfrac{x^2}{z^2}+\dfrac{z^2}{16x^2}>=2\cdot\sqrt{\dfrac{x^2}{z^2}\cdot\dfrac{z^2}{16x^2}}=\dfrac{1}{2}\)
\(\dfrac{y^2}{z^2}+\dfrac{z^2}{16y^2}>=\dfrac{1}{2}\)
và \(\dfrac{1}{x^2}+\dfrac{1}{y^2}>=\dfrac{2}{xy}>=\dfrac{2}{\left(\dfrac{x+y}{2}\right)^2}=\dfrac{8}{\left(x+y\right)^2}\)
=>\(\dfrac{15}{16}z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)>=\dfrac{15}{16}z^2\cdot\dfrac{8}{\left(x+y\right)^2}=\dfrac{15}{2}\left(\dfrac{z}{x+y}\right)^2=\dfrac{15}{2}\)
=>VT>=5+1/2+1/2+15/2=27/2
Cho x, y, z > 0 và xyz=1. CMR :
\(\dfrac{x^2}{1+y}+\dfrac{y^2}{1+z}+\dfrac{z^2}{1+z}\ge\dfrac{3}{2}\)
Đề sai nhé, \(\dfrac{z^2}{x+1}\) mới đúng nha
\(\dfrac{x^2}{y+1}+\dfrac{y^2}{z+1}+\dfrac{z^2}{x+1}\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+3}\left(\text{Svácxơ}\right)\)
\(\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
Ta có: \(x+y+z\ge3\sqrt[3]{xyz}=3\)
\(\Rightarrow x+y+z+3\le2\left(x+y+z\right)\)
Áp dụng BĐT Cauchy:
\(\sqrt{\dfrac{x}{y+z}}+\sqrt{\dfrac{y}{z+x}}+\sqrt{\dfrac{z}{x+y}}\)
\(=\dfrac{x}{\sqrt{x\left(y+z\right)}}+\dfrac{y}{\sqrt{y\left(z+x\right)}}+\dfrac{z}{\sqrt{z\left(x+y\right)}}\)
\(\ge\dfrac{x}{\dfrac{x+y+z}{2}}+\dfrac{y}{\dfrac{x+y+z}{2}}+\dfrac{z}{\dfrac{x+y+z}{2}}\)
\(=\dfrac{2x}{x+y+z}+\dfrac{2y}{x+y+z}+\dfrac{2z}{x+y+z}\)
\(=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
Dấu "=" không xảy ra nên \(\sqrt{\dfrac{x}{y+z}}+\sqrt{\dfrac{y}{z+x}}+\sqrt{\dfrac{z}{x+y}}>2\)