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Question Number 126144 by Ndala last updated on 17/Dec/20

if: a+b=c  Prove that: a^(2/3) +b^(2/3) >c^(2/3)   .  Help me, please!

$${if}:\:{a}+{b}={c} \\ $$ $${Prove}\:{that}:\:{a}^{\frac{\mathrm{2}}{\mathrm{3}}} +{b}^{\frac{\mathrm{2}}{\mathrm{3}}} >{c}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$ $$. \\ $$ $$\mathrm{H}{elp}\:{me},\:{please}! \\ $$

Answered by MJS_new last updated on 18/Dec/20

a^(2/3) +b^(2/3) >(a+b)^(2/3)   (a^(2/3) +b^(2/3) )^3 >(a+b)^2   a^2 +3a^(4/3) b^(2/3) +3a^(2/3) b^(4/3) +b^2 >a^2 +2ab+b^2   3a^(2/3) b^(2/3) (a^(2/3) +b^(2/3) )>2ab  3(a^(2/3) +b^(2/3) )>2a^(1/3) b^(1/3)   a^(2/3) −(2/3)a^(1/3) b^(1/3) +b^(2/3) >0  x^2 −(2/3)xy+y^2 >0 true for x, y ∈R\{0}

$${a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} >\left({a}+{b}\right)^{\mathrm{2}/\mathrm{3}} \\ $$ $$\left({a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} \right)^{\mathrm{3}} >\left({a}+{b}\right)^{\mathrm{2}} \\ $$ $${a}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{4}/\mathrm{3}} {b}^{\mathrm{2}/\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}/\mathrm{3}} {b}^{\mathrm{4}/\mathrm{3}} +{b}^{\mathrm{2}} >{a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} \\ $$ $$\mathrm{3}{a}^{\mathrm{2}/\mathrm{3}} {b}^{\mathrm{2}/\mathrm{3}} \left({a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} \right)>\mathrm{2}{ab} \\ $$ $$\mathrm{3}\left({a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} \right)>\mathrm{2}{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} \\ $$ $${a}^{\mathrm{2}/\mathrm{3}} −\frac{\mathrm{2}}{\mathrm{3}}{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} >\mathrm{0} \\ $$ $${x}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}{xy}+{y}^{\mathrm{2}} >\mathrm{0}\:\mathrm{true}\:\mathrm{for}\:{x},\:{y}\:\in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$

Commented byNdala last updated on 21/Dec/20

Thanks!

$$\mathrm{Thanks}! \\ $$

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