Question

Cho a, b, c > 1. Chứng minh:

a) \(\dfrac{a^2}{a-1}+\dfrac{b^2}{b-1}\ge8\)

b) \(\dfrac{a}{\sqrt{b}-1}+\dfrac{b}{\sqrt{c}-1}+\dfrac{c}{\sqrt{a}-1}\ge12\)

Answer 1

a) Áp dụng bất đẳng thức Cauchy Shwarz dạng Engel, ta có:

\(A=\dfrac{x^2}{x-1}+\dfrac{y^2}{y-1}\)

\(\ge\dfrac{\left(x+y\right)^2}{x+y-2}\)

Đặt \(x+y=a\left(a>0\right)\)

\(\Rightarrow A\ge\dfrac{a^2}{a-2}\)

\(=\dfrac{8\left(a-2\right)+\left(a^2-8a+16\right)}{a-2}\)

\(=8+\dfrac{\left(a-4\right)^2}{a-2}\ge8\)

Dấu "=" xảy ra khi \(x=y=2\)

Answer 2

b) Áp dụng bất đẳng thức Cauchy Shwarz dạng Engel, ta có:

\(A=\dfrac{x}{\sqrt{y}-1}+\dfrac{y}{\sqrt{z}-1}+\dfrac{z}{\sqrt{x}-1}\)

\(\ge\dfrac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{\sqrt{x}+\sqrt{y}+\sqrt{z}-3}\)

Đặt \(\sqrt{x}+\sqrt{y}+\sqrt{z}=a\left(a>0\right)\)

\(\Rightarrow A\ge\dfrac{a^2}{a-3}\)

\(=\dfrac{12\left(a-3\right)+\left(a^2-12a+36\right)}{a-3}\)

\(=12+\dfrac{\left(a-6\right)^2}{a-3}\ge12\)

Dấu "=" xảy ra khi x = y = z = 2