Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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 byYozzii 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 byRasheed 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.

$$\mathrm{I}\:\mathrm{admit}\:\mathrm{that}\:\mathrm{your}\:\mathrm{comment}\:\mathrm{on}\:\mathrm{Q}#\mathrm{5380},\:\mathrm{in}\:\mathrm{which}\:\mathrm{you}\:\mathrm{reached} \\ $$ $$\mathrm{accidently}\:\mathrm{at}\:\mathrm{the}\:\mathrm{result}\:\mathrm{given}\:\mathrm{in}\:\mathrm{this}\:\mathrm{question}\:\mathrm{and} \\ $$ $$\mathrm{whose}\:\mathrm{screenshot}\:\mathrm{is}\:\mathrm{given}\:\mathrm{here}\left(\mathrm{sorry}\:\mathrm{that}\:\mathrm{I}\:\mathrm{posted}\:\mathrm{it}\:\right. \\ $$ $$\left.\mathrm{without}\:\mathrm{reference}\right)\:\mathrm{is}\:\mathrm{not}\:\mathrm{direct}\:\mathrm{proof}\:\mathrm{of}\:\mathrm{your}\:\mathrm{question} \\ $$ $$\mathrm{but}\:\mathrm{it}\:\:\mathrm{may}\:\mathrm{help}\:\mathrm{to}\:\mathrm{make}\:\mathrm{proof}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com