If-a-b-R-Then-a-2-b-2-ab-a-4-b-4-2-

Question Number 205423 by hardmath last updated on 20/Mar/24

If

a, b \in R

Then :

a^{2} + b^{2} ⩾ ab + \sqrt{\frac{a^{4} + b^{4}}{2}}

Answered by Berbere last updated on 21/Mar/24

\Leftrightarrow {(a^{2} + b^{2} - a b)}^{2} ⩾ \frac{a^{4} + b^{4}}{2}

\Leftrightarrow \frac{a^{4} + b^{4}}{2} + 3 a^{2} b^{2} - 2 a b (a^{2} + b^{2}) ⩾ 0

a^{4} + b^{4} = {(a^{2} + b^{2})}^{2} - 2 a^{2} b^{2}

\Leftrightarrow \frac{{(a^{2} + b^{2})}^{2} - 2 a^{2} b^{2}}{2} + 3 a^{2} b^{2} - 2 a b (a^{2} + b^{2}) ⩾ 0

\Leftrightarrow {(a^{2} + b^{2})}^{2} + {(2 a b)}^{2} - 2 (2 a b) (a^{2} + b^{2}) ⩾ 0

\Leftrightarrow {(a^{2} + b^{2} - 2 a b)}^{2} ⩾ 0 T r u e

Commented by Skabetix last updated on 22/Mar/24

comment passer de

{(a^{2} + b^{2} - ab)}^{2} ⩾ \frac{a^{4} + b^{4}}{2}

a \frac{a^{4} + b^{4}}{2} + {3 a}^{2} b^{2} - 2 ab (a^{2} + b^{2}) ⩾ 0 ?

Commented by Berbere last updated on 24/Mar/24

s q u a r

a^{4} + b^{4} + 2 a^{2} b^{2} + a^{2} b^{2} - 2 a b (a^{2} + b^{2}) ⩾ \frac{a^{4} + b^{4}}{2}

\Rightarrow a^{4} + b^{4} - \frac{a^{4} + b^{4}}{2} + 3 a^{2} b^{2} - 2 a b (a^{2} + b^{2}) ⩾ 0

\frac{a^{4} + b^{4}}{2} + 3 a^{2} b^{2} - 2 a b (a^{2} + b^{2}) ⩾ 0

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