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Đặt \(\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)=\left(a;b;c\right)\)
BĐT cần chứng minh: \(\frac{a+b}{c^2}+\frac{b+c}{a^2}+\frac{c+a}{b^2}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(VT=a\left(\frac{1}{b^2}+\frac{1}{c^2}\right)+b\left(\frac{1}{a^2}+\frac{1}{c^2}\right)+c\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\ge2\left(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\right)\)
Mà: \(\frac{a}{bc}+\frac{c}{ab}\ge\frac{2}{b}\) ; \(\frac{a}{bc}+\frac{b}{ac}\ge\frac{2}{c}\) ; \(\frac{c}{ab}+\frac{b}{ac}\ge\frac{2}{a}\)
\(\Rightarrow2\left(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow VT\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\) (đpcm)
\(1=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{z}\right)+\frac{1}{2}\left(\frac{y}{z}+\frac{z}{x}\right)+\frac{1}{2}\left(\frac{z}{x}+\frac{x}{y}\right)\)
\(\ge\sqrt{\frac{x}{y}.\frac{y}{z}}+\sqrt{\frac{y}{z}.\frac{z}{x}}+\sqrt{\frac{z}{x}.\frac{x}{y}}=VP\) (rút gọn lại thôi:v)
Đặt \(a=\sqrt{x},b=\sqrt{y},c=\sqrt{z}\left(a,b,c>0\right)\)
Khi đó
\(P=\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)và \(a^2+b^2+c^2\ge3\)
<=>\(P=\frac{a^4}{a^2b}+\frac{b^4}{cb^2}+\frac{c^4}{ac^2}\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2b+cb^2+ac^2}\)(bất đẳng thức cosi schwaz)
Ta có
\(\left(a+b+c\right)\left(a^2+b^2+c^2\right)=\left(a^3+b^2a\right)+\left(b^3+bc^2\right)+\left(c^3+ca^2\right)+\left(a^2b+b^2c+c^2a\right)\)
\(\ge3\left(a^2b+b^2c+c^2a\right)\)
=> \(a^2b+b^2c+c^2a\le\frac{1}{3}\left(a+b+c\right)\left(a^2+b^2+c^2\right)\le\frac{\sqrt{3}}{3}\sqrt{\left(a^2+b^2+c^2\right)^3}\)
Khi đó
\(P\ge\sqrt{3}.\frac{\left(a^2+b^2+c^2\right)^2}{\sqrt{\left(a^2+b^2+c^2\right)^3}}=\sqrt{3\left(a^2+b^2+c^2\right)}\ge3\)(ĐPCM)
Dấu bằng xảy ra khi a=b=c=1 => x=y=z=1
\(\sqrt{x\left(y+z\right)}\le\frac{x+y+z}{2}\)( Cauchy)
\(\Rightarrow\sqrt{\frac{x}{y+z}}=\frac{x}{\sqrt{x\left(y+z\right)}}\le\frac{x}{\frac{x+y+z}{2}}=\frac{2x}{x+y+z}\)
Chứng minh tương tự:
\(\sqrt{\frac{y}{x+z}}\le\frac{2y}{x+y+z};\sqrt{\frac{z}{x+y}}\le\frac{2z}{x+y+z}\)
Cộng theo vế suy ra đocn. Dấu "=" ko xảy ra
Ta có: \(x+y+z=xyz\Rightarrow x=\frac{x+y+z}{yz}\Rightarrow x^2=\frac{x^2+xy+xz}{yz}\Rightarrow x^2+1=\frac{\left(x+y\right)\left(x+z\right)}{yz}\)\(\Rightarrow\sqrt{x^2+1}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{yz}}\le\frac{\frac{x+y}{y}+\frac{x+z}{z}}{2}=1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)\(\Rightarrow\frac{1+\sqrt{1+x^2}}{x}\le\frac{2+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)}{x}=\frac{2}{x}+\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)
Tương tự: \(\frac{1+\sqrt{1+y^2}}{y}\le\frac{2}{y}+\frac{1}{2}\left(\frac{1}{z}+\frac{1}{x}\right)\); \(\frac{1+\sqrt{1+z^2}}{z}\le\frac{2}{z}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Cộng theo vế ba bất đẳng thức trên, ta được: \(\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3.\frac{xy+yz+zx}{xyz}\)\(\le3.\frac{\frac{\left(x+y+z\right)^2}{3}}{xyz}=\frac{\left(x+y+z\right)^2}{xyz}=\frac{\left(xyz\right)^2}{xyz}=xyz\)
Đẳng thức xảy ra khi \(x=y=z=\sqrt{3}\)
ta có 3x + yz = x2 + xy + yz + zx = (x+y)(x+z)
do đó:
\(\frac{x}{x+\sqrt{3x+yz}}=\frac{x\left(\sqrt{x^2+xy+yz+zx}-x\right)}{\left(\sqrt{x^2+xy+yz+zx}+x\right)\left(\sqrt{x^2+xy+yz+zx}-x\right)}\)
= \(\frac{x\left(\sqrt{\left(x+y\right)\left(x+z\right)}-x\right)}{xy+yz+zx}\le\frac{x\left(\frac{x+y+x+z}{2}-x\right)}{xy+yz+zx}\)\(\le\frac{x\left(y+z\right)}{2\left(xy+yz+zx\right)}\)
tương tự với 2 số hạng còn lại nên ta được: P\(\le\)1. đpcm
