a) Sửa đề :

\(x^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)

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

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

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

\(x^4=\left(a+b\right)\left[\left(a^3+2a^2b+ab^2\right)+\left(a^2b+2ab^2+b^3\right)\right]\)

\(x^4=\left(a+b\right)\left[a\left(a^2+2ab+b^2\right)+b\left(a^2+2ab+b^2\right)\right]\)

\(x^4=\left(a+b\right)^2\left(a+2ab+b^2\right)\)

\(x^4=\left(a+b\right)^4\)

b) Sửa đề:

 \(x^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\)

\(x^5=\left(a^5+4a^4b+6a^3b^2+4a^2b^3+ab^4\right)+\left(a^4b+4a^3b^2+6a^2b+4ab^4+b^5\right)\)

\(x^5=a\left(a^4+4a^3b+6a^2b^2+4ab^3+b^4\right)+b\left(a^4+4a^3b+6a^2b^2+4ab^3+b^4\right)\)

\(x^5=\left(a+b\right)\left(a^4+4a^3b+6a^2b^2+4ab^3+b^4\right)\)

\(x^5=\left(a+b\right)\left[\left(a^4+3a^3b+3a^2b^2+ab^3\right)+\left(a^3b+3a^2b^2++3ab^3+b^4\right)\right]\)

\(x^5=\left(a+b\right)\left[a\left(a^3+3a^2b+3ab^2+b^3\right)+b\left(a^3+3a^2b+3ab^2+b^3\right)\right]\)

\(x^5=\left(a+b\right)^2\left(a^3+3a^2b+3ab^2+b^3\right)\)

\(x^5=\left(a+b\right)^2\left[\left(a^3+2a^2b+ab^2\right)+\left(a^2b+2ab^2+b^3\right)\right]\)

\(x^5=\left(a+b\right)^2\left[a\left(a^2+2ab+b^2\right)+b\left(a^2+2ab+b^2\right)\right]\)

\(x^5=\left(a+b\right)^3\left(a^2+2ab+b^2\right)\)

\(x^5=\left(a+b\right)^5\)

Bạn có thể tự tóm tắt lại

\(\frac{a^4+b^4}{2}\ge ab^3+a^3b-a^2b^2\)

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

\(\Leftrightarrow\left(a^2+b^2\right)^2-2ab\left(a^2+b^2\right)\ge\left(a^2+b^2\right).2\sqrt{a^2.b^2}-2ab\left(a^2+b^2\right)=0\)( luôn đúng )

vì BĐT cuối luôn đúng nên BĐT đã cho đúng \(\Leftrightarrow a=b\)

\(a^4+b^4\ge2a^3b+2ab^3-2a^2b^2\)

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

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

\(\Rightarrow\)Điều phải chứng minh

4 + b 4 ≥ 2a 3b + 2ab 3 − 2a 2b 2

⇔ a 4 − 2a 3b + a 2b 2 + b 4 − 2ab 3 + a 2b 2 ≥ 0

⇔ a 2 − ab 2 + b 2 − ab 2 ≥ 0 (đúng)

⇒Điều phải chứng minh

 chúc cậu hok tốt @_@

Lời giải:
Áp dụng BĐT Cô-si:
a^3+2b^3=a^3+b^3+b^3\geq 3\sqrt[3]{a^3b^6}=3ab^2$

$a^3+1+1\geq 3a$

$b^3+1+1\geq 3b$

Cộng theo vế các BĐT trên:

$a^3+2b^3+(a^3+2)+2(b^3+2)\geq 3ab^2+3a+6b$

$\Leftrightarrow 2(a^3+2b^3)+6\geq 3(ab^2+a+2b)=3.4=12$

$\Rightarrow a^3+2b^3\geq (12-6):2=3$

Ta có đpcm.

Dấu "=" xảy ra khi $a=b=1$

Ta có : \(\dfrac{4a-3b}{2}=\dfrac{5b-4c}{3}=\dfrac{3c-5a}{4}\)

\(\Leftrightarrow\dfrac{20a-15b}{10}=\dfrac{15b-12c}{9}=\dfrac{12c-20a}{16}=\dfrac{20a-15b+15b-12c+12c-20a}{10+9+16}=0\)\(\Leftrightarrow\left\{{}\begin{matrix}4a-3b=0\\5b-4c=0\\3c-5a=0\end{matrix}\right.\) 

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{3}=\dfrac{b}{4}\\\dfrac{b}{4}=\dfrac{c}{5}\\\dfrac{c}{5}=\dfrac{a}{3}\end{matrix}\right.\Leftrightarrow\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{5}\)