\(S=\dfrac{1}{a^3+b^3}+\dfrac{\dfrac{9}{4}}{3a^2b}+\dfrac{\dfrac{9}{4}}{3ab^2}+\dfrac{1}{4ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

Áp dụng bđt Cauchy-Schwarz dạng Engel có:

\(S\ge\dfrac{\left(1+\dfrac{3}{2}+\dfrac{3}{2}\right)^2}{a^3+3a^2b+3ab^2+b^3}+\dfrac{1}{4ab}.\dfrac{4}{a+b}\)

\(\Leftrightarrow S\ge\dfrac{16}{\left(a+b\right)^3}+\dfrac{1}{\left(a+b\right)^2}.\dfrac{4}{a+b}\)

\(\Leftrightarrow S\ge\dfrac{16}{1}+\dfrac{1}{1}.\dfrac{4}{1}=20\)

Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)

Vậy GTNN của \(S=20\) khi \(a=b=\dfrac{1}{2}\)

\(A=a\left(a^2+2b\right)+b\left(b^2-a\right)=a^3+2ab+b^3-ab\)

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

\(=1\cdot\left(a^2-ab+b^2\right)+ab=a^2-ab+b^2+ab\)

\(=a^2+b^2\)

\(a^2+b^2\ge0\Rightarrow A\ge0\)

A=a³+2ab+b³-ab

A=(a+b)(a²-ab+b²)+ab

A=a²+b²

Áp dg BDT cosi ta co 

a²+b²>=2ab

Dấu = xảy ra khi a=b

=>A_min=2ab <=> a=b=0,5

=>a=0,5

trả lời 

a=0,5

chúc bn 

hc tốt

Chỉ làm được 1 tý thôi:

\(a+b+1=8ab\Rightarrow\frac{a+b+1}{ab}=\frac{8ab}{ab}\)

                                   \(\Leftrightarrow\frac{1}{b}+\frac{1}{a}+\frac{1}{ab}=8.\)

Đáp án là 8 á. xảy ra khi a=b=\(\frac{1}{2}\) nhưng mình k biết cách làm.

\(A=a\left(a^2+2b\right)+b\left(b^2-a\right)=a^3+2ab+b^3-ab\)

\(=\left(a^3+b^3\right)+ab=\left(a+b\right)\left(a^2-ab+b^2\right)+ab=1\left(a^2-ab+b^2\right)-ab\)

\(=a^2-ab+b^2-ab=a^2-2ab+b^2=\left(a-b\right)^2>=0\)

dấu = xảy ra khi a=b

vậy min A là 0 khi a=b

\(A=2\left(a^2+b^2\right)=2\left[\left(b+1\right)^2+b^2\right]=2\left(2b^2+2b+1\right)=4\left[b^2+b+\dfrac{1}{4}\right]+1=4\left(b+\dfrac{1}{2}\right)^2+1\ge1\)

 " = " \(\Leftrightarrow b=-\dfrac{1}{2};a=\dfrac{1}{2}\)

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