Question

CMR nếu a, b\(\in N^{ }\)( a,b # 0) và a + b = 1 thì:

\(\left(a+\dfrac{1}{b}\right)^2+\left(b+\dfrac{1}{a}\right)^2\ge\dfrac{25}{2}\)

Các bn júp mk lm bài này.

Mk cảm ơn trước nha!!!

Answer 1

cach khac\(\left(a+\dfrac{1}{b}\right)^2+\left(b+\dfrac{1}{a}\right)^2\ge\dfrac{1}{2}\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2\ge\dfrac{1}{2}\left(a+b+\dfrac{4}{a+b}\right)^2=\dfrac{25}{2}\)

Answer 2

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\left(1^2+1^2\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2=1\)

\(\Rightarrow2\left(a^2+b^2\right)\ge1\Rightarrow a^2+b^2\ge\dfrac{1}{2}\)

Áp dụng BĐT Holder ta có:

\(\left(a+b\right)\left(a+b\right)\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\ge\left(1+1\right)^3=8\)

Lại có:

\(\left(a+\dfrac{1}{b}\right)^2+\left(b+\dfrac{1}{a}\right)^2=4+a^2+b^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}\ge4+\dfrac{1}{2}+8=\dfrac{25}{2}\)