cho a,b là các số thực dương tm \(a^3+b^3+6ab\le\) 8
cmr \(P=a+2b+\frac{2}{a}+\frac{3}{b}\ge8\)
bl
ta có \(8\ge a^3+b^3+6ab\ge ab\left(a+b\right)+6ab\ge ab\left(a+b+1+1+1+1+1+1\right)\ge8ab\sqrt[8]{ab}\)
suy ra ab\(\le1\)
mà P=\(a+b+b+\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}\ge8.\sqrt[8]{a.b.b.\frac{1}{a^2}.\frac{1}{b^3}}=8\sqrt[8]{\frac{1}{ab}}\ge8\)
dau = sảy ra khi a=b=1
tks bn