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

$$\mathrm{If}$$$$\mathrm{a},\mathrm{b}\in \mathbb{R}$$$$\mathrm{Then}:$$$${\mathrm{a}}^2+{\mathrm{b}}^2⩾\mathrm{ab}+\sqrt{\frac{{\mathrm{a}}^4+{\mathrm{b}}^4}{2}}$$

Answered by Berbere last updated on 21/Mar/24

$$\iff {(a^2+b^2-ab)}^2⩾\frac{a^4+b^4}{2}$$$$\iff \frac{a^4+b^4}{2}+3a^2{b}^2-2ab(a^2+b^2)⩾0$$$$a^4+b^4={(a^2+b^2)}^2-2a^2{b}^2$$$$\iff \frac{{(a^2+b^2)}^2-2a^2{b}^2}{2}+3a^2{b}^2-2ab(a^2+b^2)⩾0$$$$\iff {(a^2+b^2)}^2+{\left(2ab\right)}^2-2\left(2ab\right)(a^2+b^2)⩾0$$$$\iff {(a^2+b^2-2ab)}^2⩾0True$$$$$$$$$$

Commented by Skabetix last updated on 22/Mar/24

$$\mathrm{comment}\mathrm{passer}\mathrm{de}$$$${({\mathrm{a}}^2+{\mathrm{b}}^2-\mathrm{ab})}^2⩾\frac{{\mathrm{a}}^4+{\mathrm{b}}^4}{2}$$$$\mathrm{a}\frac{{\mathrm{a}}^4+{\mathrm{b}}^4}{2}+{3\mathrm{a}}^2{\mathrm{b}}^2-2\mathrm{ab}({\mathrm{a}}^2+{\mathrm{b}}^2)⩾0?$$

Commented by Berbere last updated on 24/Mar/24

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

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