Question Number 71633 by mind is power last updated on 18/Oct/19
$$\mathrm{let}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\in\mathrm{IR}_{+} \\ $$$$\mathrm{show}\:\mathrm{that}\:\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}+\mathrm{c}\right)\geqslant\mathrm{2}\sqrt{\mathrm{abc}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)} \\ $$$$ \\ $$
Answered by MJS last updated on 18/Oct/19
$${a}>\mathrm{0}\wedge{b}>\mathrm{0}\wedge{c}>\mathrm{0}\:\Rightarrow\:\left({a}+{b}\right)\left({a}+{c}\right)>\mathrm{0}\wedge{abc}\left({a}+{b}+{c}\right)>\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{are}\:\mathrm{allowed}\:\mathrm{to}\:\mathrm{square} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} \left({a}+{c}\right)^{\mathrm{2}} \geqslant\mathrm{4}{abc}\left({a}+{b}+{c}\right) \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} \left({a}+{c}\right)^{\mathrm{2}} −\mathrm{4}{abc}\left({a}+{b}+{c}\right)\geqslant\mathrm{0} \\ $$$$\left({a}−{b}\right)^{\mathrm{2}} {c}^{\mathrm{2}} +\mathrm{2}{a}\left({a}−{b}\right)\left({a}+{b}\right){c}+{a}^{\mathrm{2}} \left({a}+{b}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\left(\left({a}−{b}\right){c}+{a}\left({a}+{b}\right)\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\mathrm{true}\:\forall\:{a},\:{b},\:{c}\:\in\mathbb{R}^{+} \\ $$
Commented by mind is power last updated on 18/Oct/19
$$\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}+\mathrm{c}\right)\geqslant\mathrm{4abc}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\:\mathrm{y}'\mathrm{re}\:\mathrm{methode}\:\mathrm{worck}\:\mathrm{nice}\:\mathrm{sir} \\ $$
Commented by MJS last updated on 18/Oct/19
$$\mathrm{corrected},\:\mathrm{it}\:\mathrm{was}\:\mathrm{just}\:\mathrm{a}\:\mathrm{typo} \\ $$