Proof of one inequality a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}

Pamela Meyer 2022-01-07 Answered
Proof of one inequality
\(\displaystyle{a}+{b}+{c}\leq{\frac{{{a}^{{3}}}}{{{b}{c}}}}+{\frac{{{b}^{{3}}}}{{{c}{a}}}}+{\frac{{{c}^{{3}}}}{{{a}{b}}}}\)

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Expert Answer

Neil Dismukes
Answered 2022-01-08 Author has 3487 answers
Using Cauchy-Schwarz Inequality twice:
\(\displaystyle{a}^{{4}}+{b}^{{4}}+{c}^{{4}}\geq{a}^{{2}}{b}^{{2}}+{b}^{{2}}{c}^{{2}}+{c}^{{2}}{a}^{{2}}\geq{a}{b}^{{2}}{c}+{b}{a}^{{2}}{c}+{a}{c}^{{2}}{b}={a}{b}{c}{\left({a}+{b}+{c}\right)}\)
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Dawn Neal
Answered 2022-01-09 Author has 3373 answers
I have come up with an answer with myself. Using CS inequality
\(\displaystyle{\left({a}^{{4}}+{b}^{{4}}+{c}^{{4}}\right)}{\left({1}+{1}+{1}\right)}\geq{\left({a}^{{2}}+{b}^{{2}}+{c}^{{2}}\right)}^{{2}}\)
\(\displaystyle{\left({a}^{{2}}+{b}^{{2}}+{c}^{{2}}\right)}{\left({1}+{1}+{1}\right)}\geq{\left({a}+{b}+{c}\right)}^{{2}}\)
Hence we have
\(\displaystyle{a}^{{4}}+{b}^{{4}}+{c}^{{4}}\geq{\frac{{{\left({a}+{b}+{c}\right)}^{{4}}}}{{{27}}}}={\left({a}+{b}+{c}\right)}{\left({\frac{{{a}+{b}+{c}}}{{{3}}}}\right)}^{{3}}\geq{a}{b}{c}{\left({a}+{b}+{c}\right)}\)
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Vasquez
Answered 2022-01-11 Author has 8850 answers

By Holder
\(\sum_{\text{cyc}}\frac{a^3}{bc}\geq\frac{(a+b+c)^3}{3(ab+ac+bc)}=\frac{(a+b+c)\cdot(a+b+c)^2}{3(ab+ac+bc)}\geq a+b+c\)

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