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1-abc-3-abc-1-a-1-b-1-c-if-a-b-c-gt-0-




Question Number 5498 by Yozzii last updated on 16/May/16
(1+abc)^3 ≥abc(1+a)(1+b)(1+c)   if a,b,c>0?
$$\left(\mathrm{1}+{abc}\right)^{\mathrm{3}} \geqslant{abc}\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)\: \\ $$$${if}\:{a},{b},{c}>\mathrm{0}? \\ $$
Answered by Rasheed Soomro last updated on 17/May/16
Commented by Yozzii last updated on 17/May/16
In attempting that, I hoped that   the inequality was true, but I couldn′t  find a proof of it. Inequality proofs like these are not  a specialty of mine.
$${In}\:{attempting}\:{that},\:{I}\:{hoped}\:{that}\: \\ $$$${the}\:{inequality}\:{was}\:{true},\:{but}\:{I}\:{couldn}'{t} \\ $$$${find}\:{a}\:{proof}\:{of}\:{it}.\:{Inequality}\:{proofs}\:{like}\:{these}\:{are}\:{not} \\ $$$${a}\:{specialty}\:{of}\:{mine}. \\ $$
Commented by Rasheed Soomro last updated on 17/May/16
I admit that your comment on Q#5380, in which you reached  accidently at the result given in this question and  whose screenshot is given here(sorry that I posted it   without reference) is not direct proof of your question  but it  may help to make proof.
$$\mathrm{I}\:\mathrm{admit}\:\mathrm{that}\:\mathrm{your}\:\mathrm{comment}\:\mathrm{on}\:\mathrm{Q}#\mathrm{5380},\:\mathrm{in}\:\mathrm{which}\:\mathrm{you}\:\mathrm{reached} \\ $$$$\mathrm{accidently}\:\mathrm{at}\:\mathrm{the}\:\mathrm{result}\:\mathrm{given}\:\mathrm{in}\:\mathrm{this}\:\mathrm{question}\:\mathrm{and} \\ $$$$\mathrm{whose}\:\mathrm{screenshot}\:\mathrm{is}\:\mathrm{given}\:\mathrm{here}\left(\mathrm{sorry}\:\mathrm{that}\:\mathrm{I}\:\mathrm{posted}\:\mathrm{it}\:\right. \\ $$$$\left.\mathrm{without}\:\mathrm{reference}\right)\:\mathrm{is}\:\mathrm{not}\:\mathrm{direct}\:\mathrm{proof}\:\mathrm{of}\:\mathrm{your}\:\mathrm{question} \\ $$$$\mathrm{but}\:\mathrm{it}\:\:\mathrm{may}\:\mathrm{help}\:\mathrm{to}\:\mathrm{make}\:\mathrm{proof}. \\ $$

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