Question

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Cho x,y,z > 0 sao cho xyz = 1 và n là số nguyên dương 

Chứng minh : \(\left(\frac{1+x}{2}\right)^n+\left(\frac{1+y}{2}\right)^n+\left(\frac{1+z}{2}\right)^n\ge3\)

Answer 1

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Ta có  : \(\frac{1+x}{2}\ge\sqrt{x}\Rightarrow\left(\frac{1+x}{2}\right)^n\ge\sqrt{x^n}\) (1)

            \(\frac{1+y}{2}\ge\sqrt{y}\Rightarrow\left(\frac{1+y}{2}\right)^n\ge\sqrt{y^n}\)(2)

            \(\frac{1+z}{2}\ge\sqrt{z}\Rightarrow\left(\frac{1+z}{2}\right)^n\ge\sqrt{z^n}\)(3) 

Từ 1,2,3 \(\Rightarrow\left(\frac{1+x}{2}\right)^n+\left(\frac{1+y}{2}\right)^n+\left(\frac{1+z}{2}\right)^n\ge\sqrt{x^n}+\sqrt{y^n}+\sqrt{z^n}\)

Áp dụng BĐT Cauchy cho 3 số ta có : 

\(\sqrt{x^n}+\sqrt{y^n}+\sqrt{z^n}\ge3^3\sqrt{\sqrt{x^n}.\sqrt{y^n}.\sqrt{z^n}}=3\)

\(\Rightarrow\left(\frac{1+x}{2}\right)^n+\left(\frac{1+y}{2}\right)^n+\left(\frac{1+z}{2}\right)^n\ge3\)

Đẳng thức xảy ra <=> x = y = z = 1