\(\left(a^{10}+b^{10}\right)\left(a^2+b^2\right)\ge\)\(\left(a^8+b^8\right)\left(a^4+b^4\right)\)

\(\Leftrightarrow a^{10}b^2+a^2b^{10}\ge a^8b^4+a^4b^8\)

\(\Leftrightarrow a^8+b^8\ge a^6b^2+a^2b^6\)

\(\Leftrightarrow\left(a^2-b^2\right)\left(a^6-b^6\right)\ge0\)

Vì a^2-b^2 va a^6-b^6 cùng dấu nên ta có điều phải chứng minh.

bn có thể giải rõ hơn ko?

Lời giải:
Áp dụng BĐT Bunhiacopxky:

$(a^4+b^4)(1+1)\geq (a^2+b^2)^2\Rightarrow a^4+b^4\geq \frac{(a^2+b^2)^2}{2}$

$(a^2+b^2)(1+1)\geq (a+b)^2\Rightarrow a^2+b^2\geq \frac{(a+b)^2}{2}$

Do đó:

$a^4+b^4\geq \frac{(a+b)^4}{8}$

$\Rightarrow 8(a^4+b^4)\geq (a+b)^4$ (đpcm)

Dấu "=" xảy ra khi $a=b$

$\Rightarrow

\(8\left(a^4+b^4\right)\ge\left(a+b\right)^4\)

\(\Rightarrow8a^4+8b^4\ge\left(a+b\right)^4\)

\(\Rightarrow8\left(a^2\right)^2+8\left(b^2\right)^2\ge\left(a+b\right)^4\)

\(\Rightarrow\left(a+b\right)^4=b^4+4ab^3+6a^2b^2+4a^3+b+a^4\)

\(\Rightarrow8\left(a^4+b^4\right)\ge\left(a+b\right)^4\)(đpcm)

P/s: dấu "=" chỉ xảy ra khi a = b = 1.

\(8(a^4+b^4)\ge\left(a+b\right)^4\)

\(\Leftrightarrow\)\(8a^4+8b^4\ge a^4+4a^3b+6a^2b^2+4ab^3+b^4\)

\(\Leftrightarrow\) \(7a^4+7b^4\ge4a^3b+6a^2b^2+4ab^3\)

\(\Leftrightarrow\)\(4a^3\left(a-b\right)+4b^3\left(b-a\right)+3\left(a^4-2a^2b^2+b^4\right)\ge0\)

\(\Leftrightarrow\) \(4\left(a^3-b^3\right)\left(a-b\right)+3\left(a^2-b^2\right)^2\ge0\)

\(\Leftrightarrow\) \(4\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\) với mọi a,b

\(\Rightarrowđpcm\)

a ) CM : \(a^4+b^4\ge a^3b+b^3a\)

Giả sử điều cần c/m là đúng

\(\Rightarrow a^4+b^4-a^3b-b^3a\ge0\)

\(\Rightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)

\(\Rightarrow\left(a^3-b^3\right)\left(a-b\right)\ge0\)

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

Ta có : \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\a^2+ab+b^2=\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}\ge0\end{matrix}\right.\)

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

\(\Rightarrow a^4+b^4-a^3b-b^3a\ge0\)

\(\Rightarrow a^4+b^4\ge a^3b+b^3a\)

\(\Rightarrow2\left(a^4+b^4\right)\ge a^4+a^3b+b^4+b^3a\)

\(\Rightarrow2\left(a^4+b^4\right)\ge\left(a+b\right)\left(a^3+b^3\right)\)

\(\left(đpcm\right)\)

b ) \(\left(a+b+c\right)\left(a^3+b^3+c^3\right)\)

\(=a^4+a^3b+a^3c+b^3a+b^4+b^3c+c^3a+c^3b+c^4\)

\(=\left(a^4+b^4+c^4\right)+\left(a^3b+b^3a\right)+\left(b^3c+c^3b\right)+\left(a^3c+c^3a\right)\)

CMTT như a ) : \(\left\{{}\begin{matrix}a^4+b^4\ge a^3b+b^3a\\b^4+c^4\ge b^3c+c^3b\\a^4+c^4\ge a^3c+c^3a\end{matrix}\right.\)

\(\Rightarrow2\left(a^4+b^4+c^4\right)\ge a^3b+b^3a+b^3c+c^3b+a^3c+c^3a\)

\(\Rightarrow3\left(a^4+b^4+c^4\right)\ge a^4+b^4+c^4+a^3b+b^3a+b^3c+c^3b+a^3c+c^3a\)

\(\Rightarrow3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\left(đpcm\right)\)

\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)

\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)

\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)

\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)

\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)

Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)

\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)

Từ ( 1 ) và ( 2 ) có đpcm

Câu a : \(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\Leftrightarrow\left(a-b\right)^2\ge0\)

a: =>2a^2+2b^2>=a^2+2ab+b^2

=>a^2-2ab+b^2>=0

=>(a-b)^2>=0(luôn đúng)

c: =>3a^2+3b^2+3c^2>=a^2+b^2+c^2+2ab+2bc+2ac

=>2a^2+2b^2+2c^2-2ab-2bc-2ac>=0

=>(a-b)^2+(b-c)^2+(a-c)^2>=0(luôn đúng)