Question

Prove: a^2+b^2/2 ≥ab

Answer 1

We know that the square of any number is greater than or equal to zero ;

So if have a number like x :

`( {x})^(2) \geqslant 0`

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Now we have a number which is

( a - b ) ;

So we have :

`( {a - b})^(2) \geqslant 0`

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Reminder :

`( {a - b})^(2) = {a}^(2) - 2ab + {b}^(2)`

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So we have :

`{a}^(2) - 2ab + {b}^(2) \geqslant 0`

Both sides plus 2ab :

`{a}^(2) + {b}^(2) \geqslant 2ab`

Divided both sides by 2 :

`\frac{ {a}^(2) + {b}^(2) }{2} \geqslant ab \\`

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And we're done.

Thanks for watching buddy good luck.

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