\(x^2+y^2\ge2\sqrt{x^2y^2}\ge2xy\)

\(x^2y^2+1\ge2\sqrt{x^2y^2.1}\ge2xy\)

\(\Rightarrow x^2+y^2+x^2.y^2+1\ge2xy+2xy=4xy\)

fuck you beast

cái đệch gì ? 

\(\frac{a}{c}+\frac{c}{a}-2=\frac{a}{c}+\frac{c}{a}-2\cdot\sqrt{\frac{a}{c}}\cdot\sqrt{\frac{c}{a}}=\left(\sqrt{\frac{a}{c}}-\sqrt{\frac{c}{a}}\right)^2\ge0\forall a,c\) >0

=>\(\frac{a}{c}+\frac{c}{a}\ge2\left(1\right)\)

\(\frac{a}{b}+\frac{b}{a}-2=\frac{a}{b}-2\cdot\sqrt{\frac{a}{b}}\cdot\sqrt{\frac{b}{a}}+\frac{b}{a}=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\ge0\forall a,b>0\)

=>\(\frac{a}{b}+\frac{b}{a}\ge2\left(2\right)\)

\(\frac{b}{c}+\frac{c}{b}-2=\frac{b}{c}-2\cdot\sqrt{\frac{b}{c}}\cdot\sqrt{\frac{c}{b}}+\frac{c}{b}=\left(\sqrt{\frac{b}{c}}-\sqrt{\frac{c}{b}}\right)^2\ge0\forall b,c>0\)

=>\(\frac{b}{c}+\frac{c}{b}\ge2\left(3\right)\)

Từ (1),(2),(3) suy ra \(\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge2+2+2\)

=>\(\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge6\forall a,b,c>0\)

=>ĐPCM

P=1+2x/(x^2+1)<= 1+(x^2+1)/(x^2+1)=2
Suy ra Pmax =2 khi x=1

Điều kiện \(a>0\)

\(A=\sqrt[4]{\frac{3}{4a}}.\sqrt[4]{\frac{4a}{3}}.x\sqrt{a-x^4}\le\sqrt[4]{\frac{3}{4a}}\left(-x^4+\sqrt{\frac{4a}{3}}x^2+a\right)\)

\(A\le\sqrt[4]{\frac{3}{4a}}\left[\frac{4a}{3}-\left(x^2-\sqrt{\frac{a}{3}}\right)^2\right]\le\frac{4a}{3}\sqrt[4]{\frac{3}{4a}}\)

Dấu "=" xảy ra khi \(x=\sqrt[4]{\frac{a}{3}}\)

=>

=>

=>

Tương tự, ta có: 

Do đó, ta có:

(ĐPCM)

Chứng minh bằng biến đổi tương đương : 

\(\frac{a+b}{2}\ge\sqrt{ab}\) \(\Leftrightarrow a+b\ge2\sqrt{ab}\Leftrightarrow a+b-2\sqrt{ab}\ge0\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge\) (luôn đúng)

Bđt cuối luôn đúng nên bđt ban đầu được chứng minh.

Dấu "=" xảy ra khi \(\sqrt{a}-\sqrt{b}=0\Leftrightarrow a=b\) (a,b không âm)