Tóm tắt :

 \(a\ge0;b\ge0\rightarrow a+b\le\sqrt{2\left(a^2+b^2\right)}\)

Ta có :

\(\left(a-b\right)^2\ge0\)

\(\Rightarrow a^2+b^2-2ab\ge0\)

\(\Rightarrow a^2+b^2\ge2ab\)

\(\Rightarrow2\left(a^2+b^2\right)\ge a^2+b^2+2ab\)

\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\Rightarrow a+b\le\sqrt{2\left(a^2+b^2\right)}\)

Vậy \(a+b\le\sqrt{2\left(a^2+b^2\right)}.\)

thanh nha

ta có : \(a^8+b^8-a^6b^2-a^2b^6\ne\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)\)

và \(a^2b^2\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)\) cũng có thể âm

\(\Rightarrow\) sai

Tóm tắt :

\(a\ge0;b\ge0\rightarrow a+b\le\sqrt{2\left(a^2+b^2\right)}\)

Ta có : 

\(\left(a-b\right)^2\ge0\)

\(\Rightarrow a^2+b^2-2ab\ge0\)

\(\Rightarrow a^2+b^2\ge2ab\)

\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\Rightarrow a+b\le\sqrt{2\left(a^2+b^2\right)}\)

Lời giải:

Áp dụng BĐT AM-GM:

$A=a^2b^2(a^2+b^2)$

$4A=2ab.2ab(a^2+b^2)\leq \left(\frac{2ab+2ab+a^2+b^2}{3}\right)^3$

$=[\frac{(a+b)^2+2ab}{3}]^3=(\frac{16+2ab}{3})^3$

Mà: 
$2ab\leq 2(\frac{a+b}{2})^2=2(\frac{4}{2})^2=8$

$\Rightarrow 4A\leq (\frac{16+8}{3})^3=512$

$\Rightarrow A\leq 128$

Dấu "=" xảy ra khi $a=b=2$

\(VT=a^2b^2\left(a^2+b^2\right)=\frac{2a^2b^2\left(a^2+b^2\right)}{2}\)

\(\le\frac{\frac{\left(a+b\right)^2}{4}}{2}\cdot\left(\frac{a^2+b^2+2ab}{4}\right)\)

\(=\frac{\frac{\left(a+b\right)^2}{4}}{2}\cdot\left(\frac{\left(a+b\right)^2}{4}\right)\)

\(\le\frac{\frac{1^2}{4}}{2}\cdot\left(\frac{1^2}{4}\right)=\frac{1}{32}\)

Dấu "=" khi \(a=b=\frac{1}{2}\)

Lời giải:
Áp dụng BĐT AM-GM:

$ab^2-a^2b=ab(b-a)\leq a(1-a)\leq (\frac{a+1-a}{2})^2=(\frac{1}{2})^2=\frac{1}{4}$

Ta có đpcm

Giá trị này đạt tại $b=1; a=\frac{1}{2}$