Trường hợp n = 2

Với mọi thực x 1 , x 2 ≥ 0 {\displaystyle x_{1},x_{2}\geq 0} , ta luôn có:

( x 1 − x 2 ) 2 ≥ 0 ⇔ x 1 − 2 x 1 x 2 + x 2 ≥ 0 ⇔ x 1 + x 2 ≥ 2 x 1 x 2 ⇔ x 1 + x 2 2 ≥ x 1 x 2 {\displaystyle ({\sqrt {x_{1}}}-{\sqrt {x_{2}}})^{2}\geq 0\Leftrightarrow x_{1}-2{\sqrt {x_{1}x_{2}}}+x_{2}\geq 0\Leftrightarrow x_{1}+x_{2}\geq 2{\sqrt {x_{1}x_{2}}}\Leftrightarrow {\frac {x_{1}+x_{2}}{2}}\geq {\sqrt {x_{1}x_{2}}}}

Trường hợp n = 2k

Giả sử

x 1 + x 2 + . . . + x k k ≥ x 1 x 2 . . . x k k {\displaystyle {\frac {x_{1}+x_{2}+...+x_{k}}{k}}\geq {\sqrt[{k}]{x_{1}x_{2}...x_{k}}}}

Ta có:

x 1 + x 2 + . . . + x k + x k + 1 + x k + 2 + . . . + x 2 k ≥ k x 1 x 2 . . . x k k + k x k + 1 x k + 2 . . . x 2 k k ( 1 ) {\displaystyle x_{1}+x_{2}+...+x_{k}+x_{k+1}+x_{k+2}+...+x_{2k}\geq k{\sqrt[{k}]{x_{1}x_{2}...x_{k}}}+k{\sqrt[{k}]{x_{k+1}x_{k+2}...x_{2k}}}(1)}

Áp dụng bất đẳng thức Côsi với trường hợp n = 2 {\displaystyle n=2} , ta lại có:

k x 1 x 2 . . . x k k + k x k + 1 x k + 2 . . . x 2 k k ≥ 2 k x 1 x 2 . . . x 2 k 2 k ( 2 ) {\displaystyle k{\sqrt[{k}]{x_{1}x_{2}...x_{k}}}+k{\sqrt[{k}]{x_{k+1}x_{k+2}...x_{2k}}}\geq 2k{\sqrt[{2k}]{x_{1}x_{2}...x_{2k}}}(2)}

Từ ( 1 ) {\displaystyle (1)} và ( 2 ) {\displaystyle (2)} , ta có được bất đẳng thức

x 1 + x 2 + . . . + x 2 k ≥ 2 k x 1 x 2 . . . x 2 k 2 k {\displaystyle x_{1}+x_{2}+...+x_{2k}\geq 2k{\sqrt[{2k}]{x_{1}x_{2}...x_{2k}}}} (đpcm)

Trường hợp n = 2k - 1

Giả sử

x 1 + x 2 + . . . + x k k ≥ x 1 x 2 . . . x k k {\displaystyle {\frac {x_{1}+x_{2}+...+x_{k}}{k}}\geq {\sqrt[{k}]{x_{1}x_{2}...x_{k}}}}

Ta có:

x 1 + x 2 + . . . + x k + x k + 1 + x k + 2 . . . + x 2 k − 1 + x 1 x 2 . . . x 2 k − 1 2 k − 1 ≥ k x 1 x 2 . . . x k k + k x k + 1 x k + 2 . . . x 2 k − 1 x 1 x 2 . . . x 2 k − 1 2 k − 1 k ( 3 ) {\displaystyle x_{1}+x_{2}+...+x_{k}+x_{k+1}+x_{k+2}...+x_{2k-1}+{\sqrt[{2k-1}]{x_{1}x_{2}...x_{2k-1}}}\geq k{\sqrt[{k}]{x_{1}x_{2}...x_{k}}}+k{\sqrt[{k}]{x_{k+1}x_{k+2}...x_{2k-1}{\sqrt[{2k-1}]{x_{1}x_{2}...x_{2k-1}}}}}(3)}

Áp dụng bất đẳng thức Côsi với trường hợp n = 2 {\displaystyle n=2} , ta lại có:

k x 1 x 2 . . . x k k + k x k + 1 x k + 2 . . . x 1 x 2 . . . x 2 k − 1 2 k − 1 k ≥ 2 k x 1 x 2 . . . x 2 k − 1 x 1 x 2 . . . x 2 k − 1 2 k − 1 2 k = 2 k x 1 x 2 . . . x 2 k − 1 2 k − 1 ( 4 ) {\displaystyle k{\sqrt[{k}]{x_{1}x_{2}...x_{k}}}+k{\sqrt[{k}]{x_{k+1}x_{k+2}...{\sqrt[{2k-1}]{x_{1}x_{2}...x_{2k-1}}}}}\geq 2k{\sqrt[{2k}]{x_{1}x_{2}...x_{2k-1}{\sqrt[{2k-1}]{x_{1}x_{2}...x_{2k-1}}}}}=2k{\sqrt[{2k-1}]{x_{1}x_{2}...x_{2k-1}}}(4)}

Từ ( 3 ) {\displaystyle (3)} và ( 4 ) {\displaystyle (4)} , ta có:

x 1 + x 2 + . . . + x k + x k + 1 + . . . + x 2 k − 1 + x 1 x 2 . . . x 2 k − 1 2 k − 1 ≥ 2 k x 1 x 2 . . . x 2 k − 1 2 k − 1 {\displaystyle x_{1}+x_{2}+...+x_{k}+x_{k+1}+...+x_{2k-1}+{\sqrt[{2k-1}]{x_{1}x_{2}...x_{2k-1}}}\geq 2k{\sqrt[{2k-1}]{x_{1}x_{2}...x_{2k-1}}}}

Cuối cùng, ta được bất đẳng thức:

x 1 + x 2 . . . + x 2 k − 1 ≥ ( 2 k − 1 ) x 1 x 2 . . . x 2 k − 1 2 k − 1 {\displaystyle x_{1}+x_{2}...+x_{2k-1}\geq (2k-1){\sqrt[{2k-1}]{x_{1}x_{2}...x_{2k-1}}}} (đpcm)