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 by Ndala last updated on 21/Dec/20
$$\mathrm{Thanks}! \\ $$