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Question Number 103310 by Quvonchbek last updated on 14/Jul/20
Commented by Quvonchbek last updated on 14/Jul/20
prove
$$\boldsymbol{{prove}} \\ $$
Answered by 1549442205 last updated on 14/Jul/20
Applying Cauchy′s inequality for three positive numbers we have: (a^3 /(b+c))+((b+c)/4)+(1/2)≥3^3 (√((a^3 /(b+c)).((b+c)/4).(1/2)))=((3a)/2) Similarly,we have: (b^3 /(c+a))+((c+a)/4)+(1/2)≥((3b)/2) (c^3 /(a+b))+((a+b)/4)+(1/2)≥((3c)/2).Adding three above inequalities we get LHS+((a+b+c)/2)+(3/2)≥((3(a+b+c))/2) ⇔LHS≥a+b+c−(3/2)(1) On the other hands, a+b+c≥3^3 (√(abc))=3(2)(due to abc=1 (bythe hypothesis)).From (1) and (2) we get (a^3 /(b+c))+(b^3 /(c+a))+(c^3 /(a+b))≥(3/2) (q.e.d) The equality ocurrs if and only if a=b=c=1 second way: Applying Cauchy−Schwarz we have: (a^3 /(b+c))+(b^3 /(c+a))+(c^3 /(a+b))⇔(a^4 /(a(b+c)))+(b^4 /(b(c+a)))+(c^4 /(c(a+b))) ≥(((a^2 +b^2 +c^2 )^2 )/(2(ab+bc+ca)))≥(((ab+bc+ca)^2 )/(2(ab+bc+ca)))=((ab+bc+ca)/2) (3) On the other hands, ab+bc+ca≥3^3 (√(ab.bc.ca))=3^3 (√((abc)^2 ))=3(4) (due to abc=1(by the hypothesis)) From(3)and (4) we get (a^3 /(b+c))+(b^3 /(c+a))+(c^3 /(a+b))≥(3/2).The equality ocurrs if and only if a=b=c=1(q.e.d)
$$\mathrm{Applying}\:\mathrm{Cauchy}'\mathrm{s}\:\mathrm{inequality}\:\mathrm{for}\: \\ $$$$\mathrm{three}\:\mathrm{positive}\:\mathrm{numbers}\:\mathrm{we}\:\mathrm{have}: \\ $$$$\frac{\mathrm{a}^{\mathrm{3}} }{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}+\mathrm{c}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\geqslant\mathrm{3}\:^{\mathrm{3}} \sqrt{\frac{\mathrm{a}^{\mathrm{3}} }{\mathrm{b}+\mathrm{c}}.\frac{\mathrm{b}+\mathrm{c}}{\mathrm{4}}.\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\mathrm{3a}}{\mathrm{2}} \\ $$$$\mathrm{Similarly},\mathrm{we}\:\mathrm{have}:\:\frac{\mathrm{b}^{\mathrm{3}} }{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}+\mathrm{a}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\geqslant\frac{\mathrm{3b}}{\mathrm{2}} \\ $$$$\frac{\mathrm{c}^{\mathrm{3}} }{\mathrm{a}+\mathrm{b}}+\frac{\mathrm{a}+\mathrm{b}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\geqslant\frac{\mathrm{3c}}{\mathrm{2}}.\mathrm{Adding}\:\mathrm{three} \\ $$$$\mathrm{above}\:\mathrm{inequalities}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{LHS}+\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}\geqslant\frac{\mathrm{3}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)}{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{LHS}\geqslant\mathrm{a}+\mathrm{b}+\mathrm{c}−\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{1}\right) \\ $$$$\mathrm{On}\:\mathrm{the}\:\mathrm{other}\:\mathrm{hands}, \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}\geqslant\mathrm{3}\:^{\mathrm{3}} \sqrt{\mathrm{abc}}=\mathrm{3}\left(\mathrm{2}\right)\left(\mathrm{due}\:\mathrm{to}\:\mathrm{abc}=\mathrm{1}\right. \\ $$$$\left.\:\left(\mathrm{bythe}\:\mathrm{hypothesis}\right)\right).\mathrm{From}\:\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right) \\ $$$$\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{get}}\:\frac{\boldsymbol{\mathrm{a}}^{\mathrm{3}} }{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}}+\frac{\boldsymbol{\mathrm{b}}^{\mathrm{3}} }{\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{a}}}+\frac{\boldsymbol{\mathrm{c}}^{\mathrm{3}} }{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}}\geqslant\frac{\mathrm{3}}{\mathrm{2}}\:\left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right) \\ $$$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{equality}}\:\boldsymbol{\mathrm{ocurrs}}\:\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{only}}\:\boldsymbol{\mathrm{if}} \\ $$$$\boldsymbol{\mathrm{a}}=\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{c}}=\mathrm{1} \\ $$$$\boldsymbol{\mathrm{second}}\:\boldsymbol{\mathrm{way}}: \\ $$$$\mathrm{Applying}\:\mathrm{Cauchy}−\mathrm{Schwarz}\:\mathrm{we}\:\mathrm{have}: \\ $$$$\frac{\boldsymbol{\mathrm{a}}^{\mathrm{3}} }{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}}+\frac{\boldsymbol{\mathrm{b}}^{\mathrm{3}} }{\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{a}}}+\frac{\boldsymbol{\mathrm{c}}^{\mathrm{3}} }{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}}\Leftrightarrow\frac{\boldsymbol{\mathrm{a}}^{\mathrm{4}} }{\mathrm{a}\left(\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)}+\frac{\boldsymbol{\mathrm{b}}^{\mathrm{4}} }{\mathrm{b}\left(\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{a}}\right)}+\frac{\boldsymbol{\mathrm{c}}^{\mathrm{4}} }{\mathrm{c}\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)} \\ $$$$\geqslant\frac{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)}\geqslant\frac{\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)}=\frac{\mathrm{ab}+\mathrm{bc}+\mathrm{ca}}{\mathrm{2}}\:\left(\mathrm{3}\right) \\ $$$$\mathrm{On}\:\mathrm{the}\:\mathrm{other}\:\mathrm{hands}, \\ $$$$\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\geqslant\mathrm{3}\:^{\mathrm{3}} \sqrt{\mathrm{ab}.\mathrm{bc}.\mathrm{ca}}=\mathrm{3}\:^{\mathrm{3}} \sqrt{\left(\mathrm{abc}\right)^{\mathrm{2}} }=\mathrm{3}\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{due}\:\mathrm{to}\:\mathrm{abc}=\mathrm{1}\left(\mathrm{by}\:\mathrm{the}\:\mathrm{hypothesis}\right)\right) \\ $$$$\mathrm{From}\left(\mathrm{3}\right)\mathrm{and}\:\left(\mathrm{4}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$$\frac{\boldsymbol{\mathrm{a}}^{\mathrm{3}} }{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}}+\frac{\boldsymbol{\mathrm{b}}^{\mathrm{3}} }{\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{a}}}+\frac{\boldsymbol{\mathrm{c}}^{\mathrm{3}} }{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}}\geqslant\frac{\mathrm{3}}{\mathrm{2}}.\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{equality}}\:\boldsymbol{\mathrm{ocurrs}} \\ $$$$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{only}}\:\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{a}}=\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{c}}=\mathrm{1}\left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right) \\ $$$$ \\ $$

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