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.

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