Question Number 126144 by Ndala last updated on 17/Dec/20 | ||
$${if}:\:{a}+{b}={c} \\ $$ $${Prove}\:{that}:\:{a}^{\frac{\mathrm{2}}{\mathrm{3}}} +{b}^{\frac{\mathrm{2}}{\mathrm{3}}} >{c}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$ $$. \\ $$ $$\mathrm{H}{elp}\:{me},\:{please}! \\ $$ | ||
Answered by MJS_new last updated on 18/Dec/20 | ||
$${a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} >\left({a}+{b}\right)^{\mathrm{2}/\mathrm{3}} \\ $$ $$\left({a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} \right)^{\mathrm{3}} >\left({a}+{b}\right)^{\mathrm{2}} \\ $$ $${a}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{4}/\mathrm{3}} {b}^{\mathrm{2}/\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}/\mathrm{3}} {b}^{\mathrm{4}/\mathrm{3}} +{b}^{\mathrm{2}} >{a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} \\ $$ $$\mathrm{3}{a}^{\mathrm{2}/\mathrm{3}} {b}^{\mathrm{2}/\mathrm{3}} \left({a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} \right)>\mathrm{2}{ab} \\ $$ $$\mathrm{3}\left({a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} \right)>\mathrm{2}{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} \\ $$ $${a}^{\mathrm{2}/\mathrm{3}} −\frac{\mathrm{2}}{\mathrm{3}}{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} >\mathrm{0} \\ $$ $${x}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}{xy}+{y}^{\mathrm{2}} >\mathrm{0}\:\mathrm{true}\:\mathrm{for}\:{x},\:{y}\:\in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$ | ||
Commented byNdala last updated on 21/Dec/20 | ||
$$\mathrm{Thanks}! \\ $$ | ||