Question

Cho a,b > 0. Cmr :

a) \(\left(1+\frac{a}{b}\right)^5+\left(1+\frac{b}{a}\right)^5\ge64\)

b) \(\frac{1}{1+3ab+a^2}+\frac{1}{1+3ab+b^2}\ge1\) với a + b = 1.

Answer 1

b) Áp dụng BĐT Cauchy-schwarz ta có:

\(\frac{1}{1+3ab+a^2}+\frac{1}{1+3ab+b^2}\ge\frac{4}{2+a^2+2ab+b^2+4ab}\)\(=\frac{4}{2+\left(a+b\right)^2+4ab}\) (1)

Dấu " = " xảy ra <=> a=b=0,5

Áp dụng BĐT AM-GM ta có:

\(4ab=4.\sqrt{ab}.\sqrt{ab}\le\frac{4.\left(a+b\right)^2}{4}=\left(a+b\right)^2=1\)(2)

Dấu " = " xảy ra <=> a=b=0,5

Từ (1) và (2)

\(\Rightarrow\frac{1}{1+3ab+a^2}+\frac{1}{1+3ab+b^2}\ge\frac{4}{2+\left(a+b\right)^2+4ab\ge}\frac{4}{3+\left(a+b\right)^2}=\frac{4}{4}=1\)

Dấu " = " xảy ra <=> a=b=0,5

Answer 2

P/s : Làm siêu tắt

Ta có :

\(\left(1+\frac{a}{b}\right)^5+\left(1+\frac{b}{a}\right)^5\ge\left(1+\frac{a}{b}\right)\left(1+\frac{b}{a}\right)\left[\left(1+\frac{a}{b}\right)^3+\left(1+\frac{b}{a}\right)^3\right]\ge\left(1+\frac{a}{b}\right)^2\left(1+\frac{b}{a}\right)^2\left(2+\frac{a}{b}+\frac{b}{a}\right)=\frac{\left(a+b\right)^2.\left(a+b\right)^2}{a^2b^2}.\left(2+\frac{a}{b}+\frac{b}{a}\right)\ge\frac{4ab.4ab}{a^2b^2}.\left(2+2\right)=16.4=64\)

( AD BĐT phụ \(x^5+y^5\ge xy\left(x^3+y^3\right);x^3+y^3\ge xy\left(x+y\right)\) và BĐT Cô - si )

Dấu " = " xảy ra \(\Leftrightarrow a=b;a,b>0\)

Answer 3

@Nguyễn Việt Lâm, @Ribi Nkok Ngok, @Khôi Bùi

Giúp vs ạ !