Prove: a^2+b^2/2 ≥ab

Question

asked Jul 18, 2021 121k views

1 Answer

Bence Pattogato · Answer 1 · 2021-07-23T03:25:03+0000

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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