Question

Bài 1 : Cho \(a\ge1;b\ge1\)

\(\dfrac{b\sqrt{a-1}+a\sqrt{b-1}}{ab}\)\(\le1\)

Bài 2 : Cho x, y thỏa mãn x + y = \(\dfrac{5}{4}\)

CMR : \(\dfrac{4}{x}+\dfrac{1}{4y}\ge5\)

Bài 3 : Cho a > 0. CMR : \(a^2+\dfrac{2}{a^3}\ge\dfrac{5\sqrt[5]{9}}{3}\)

Answer 1

1) Đặt \(\dfrac{b\sqrt{a-1}+a\sqrt{b-1}}{ab}\) là A

\(\)\(A=\dfrac{\sqrt{a-1}}{a}+\dfrac{\sqrt{b-1}}{b}\)

\(\left(\dfrac{\sqrt{a-1}}{a}\right)^2=\dfrac{a-1}{a^2}=\dfrac{1}{a}-\dfrac{1}{a^2}=\dfrac{1}{a}\left(1-\dfrac{1}{a}\right)\)

\(\Rightarrow\)\(\dfrac{\sqrt{a-1}}{a}=\sqrt{\dfrac{1}{a}\left(1-\dfrac{1}{a}\right)}\)

Tương tự: \(\dfrac{\sqrt{b-1}}{b}=\sqrt{\dfrac{1}{b}\left(\dfrac{1}{b}-1\right)}\)

Áp dụng BĐT Cauchy, ta có:

\(\sqrt{\dfrac{1}{a}\left(1-\dfrac{1}{a}\right)}\le\dfrac{\dfrac{1}{a}+\left(1-\dfrac{1}{a}\right)}{2}=\dfrac{1}{2}\)

Tương tự: \(\sqrt{\dfrac{1}{b}\left(\dfrac{1}{b}-1\right)}\le\dfrac{1}{2}\)

Cộng vế theo vế của 2 BĐT vừa chứng minh, ta được:

\(A\le1\left(đpcm\right)\)

Answer 2

Xét: \(a^2+\dfrac{2}{a^3}=\dfrac{1}{3}a^2+\dfrac{1}{3}a^2+\dfrac{1}{3}a^2+\dfrac{1}{a^3}+\dfrac{1}{a^3}\left(1\right)\)

Áp dụng BĐT Cauchy cho 5 số dương trên, ta có: \(\left(1\right)\ge5\sqrt[5]{\dfrac{1}{3}a^2.\dfrac{1}{3}a^2.\dfrac{1}{3}a^2.\dfrac{1}{a^3}.\dfrac{1}{a^3}}=5\sqrt[5]{\dfrac{1}{27}}=\dfrac{5\sqrt[5]{9}}{3}\left(đpcm\right)\)

Dấu ''='' xảy ra khi và chỉ khi \(\dfrac{1}{3}a^2=\dfrac{1}{a^3}\Leftrightarrow a=\sqrt[5]{3}\)

Answer 3

Bài 2:

Áp dụng bất đẳng thức Cauchy-Shwarz dạng Engel có:

\(\dfrac{4}{x}+\dfrac{1}{4y}=\dfrac{4^2}{4x}+\dfrac{1}{4y}\ge\dfrac{\left(4+1\right)^2}{4\left(x+y\right)}=\dfrac{25}{5}=5\)

Dấu " = " khi x = y = \(\dfrac{5}{8}\)