Question

Cho hai số a;b>0 thỏa mãn \(a+\dfrac{1}{b}=1\) .Chứng minh: \(\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2\)≥\(\dfrac{25}{2}\)

Answer 1

Lời giải:

$a+\frac{1}{b}=1\Rightarrow b=\frac{1}{1-a}$

Khi đó:

$A=(a+\frac{1}{a})^2+(b+\frac{1}{b})^2=a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2}+4$

$=(1-a)^2+\frac{1}{(1-a)^2}+a^2+\frac{1}{a^2}+4$

Áp dụng BĐT AM-GM:

$A=[\frac{1}{(1-a)^2}+\frac{1}{a^2}]+[(1-a)^2+a^2]$

$\geq \frac{2}{a(1-a)}+2a(1-a)+4$

$=2a(1-a)+\frac{1}{8a(1-a)}+\frac{15}{8a(1-a)}+4$

\(\geq 2\sqrt{2a(1-a).\frac{1}{8a(1-a)}}+\frac{15}{8.\left(\frac{a+1-a}{2}\right)^2}+4\)

\(=2\sqrt{\frac{1}{4}}+\frac{15}{2}+4=\frac{25}{2}\)

Ta có đpcm

Dấu "=" xảy ra khi $a=\frac{1}{2}; b=2$