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Question Number 71633 by mind is power last updated on 18/Oct/19

$$\mathrm{let}\mathrm{a},\mathrm{b},\mathrm{c}\in {\mathrm{IR}}_{+}$$$$\mathrm{show}\mathrm{that}(\mathrm{a}+\mathrm{b})(\mathrm{a}+\mathrm{c})⩾2\sqrt{\mathrm{abc}(\mathrm{a}+\mathrm{b}+\mathrm{c})}$$$$$$

Answered by MJS last updated on 18/Oct/19

$$a>0\wedge b>0\wedge c>0\Rightarrow (a+b)(a+c)>0\wedge abc(a+b+c)>0$$$$\Rightarrow \mathrm{we}\mathrm{are}\mathrm{allowed}\mathrm{to}\mathrm{square}$$$${(a+b)}^2{(a+c)}^2⩾4abc(a+b+c)$$$${(a+b)}^2{(a+c)}^2-4abc(a+b+c)⩾0$$$${(a-b)}^2{c}^2+2a(a-b)(a+b)c+a^2{(a+b)}^2⩾0$$$${((a-b)c+a(a+b))}^2⩾0$$$$\mathrm{true}\mathrm{\forall }a,b,c\in {\mathbb{R}}^{+}$$

Commented by mind is power last updated on 18/Oct/19

$$(\mathrm{a}+\mathrm{b})(\mathrm{a}+\mathrm{c})⩾4\mathrm{abc}(\mathrm{a}+\mathrm{b}+\mathrm{c}){\mathrm{y}}^{\prime }\mathrm{re}\mathrm{methode}\mathrm{worck}\mathrm{nice}\mathrm{sir}$$

Commented by MJS last updated on 18/Oct/19

$$\mathrm{corrected},\mathrm{it}\mathrm{was}\mathrm{just}\mathrm{a}\mathrm{typo}$$

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