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Question Number 98570 by HamraboyevFarruxjon last updated on 14/Jun/20

a,b,c>0 prove: (a/(√(a^2 +8bc)))+(b/(√(b^2 +8ac)))+(c/(√(c^2 +8ab)))≥1 help please...

$$\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}>\mathrm{0}\:\:\:\:\:\:\:\boldsymbol{{prove}}: \\ $$ $$\frac{\boldsymbol{{a}}}{\sqrt{\boldsymbol{{a}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{{bc}}}}+\frac{\boldsymbol{{b}}}{\sqrt{\boldsymbol{{b}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{{ac}}}}+\frac{\boldsymbol{{c}}}{\sqrt{\boldsymbol{{c}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{{ab}}}}\geqslant\mathrm{1} \\ $$ $$\boldsymbol{{help}}\:\boldsymbol{{please}}... \\ $$

Commented byMJS last updated on 14/Jun/20

extremes where a=b=c here this leads to 3(a/(√(a^2 +8a^2 )))=1 for a>0 ⇒ min (lhs) =1 ⇒ proven

$$\mathrm{extremes}\:\mathrm{where}\:{a}={b}={c} \\ $$ $$\mathrm{here}\:\mathrm{this}\:\mathrm{leads}\:\mathrm{to} \\ $$ $$\mathrm{3}\frac{{a}}{\sqrt{{a}^{\mathrm{2}} +\mathrm{8}{a}^{\mathrm{2}} }}=\mathrm{1}\:\mathrm{for}\:{a}>\mathrm{0} \\ $$ $$\Rightarrow\:\mathrm{min}\:\left(\mathrm{lhs}\right)\:=\mathrm{1}\:\Rightarrow\:\mathrm{proven} \\ $$

Answered by 1549442205 last updated on 15/Jun/20

Applying the Cauchi−Schwartz we have we have P=(a/(√(a^2 +8bc)))+(b/(√(b^2 +8ca)))+(c/(√(c^2 +8ab)))=(a^2 /(a(√(a^2 +8bc))))+(b^2 /(b(√(b^2 +8ca))))+(c^2 /(c(√(c^2 +8ab))))≥(((a+b+c)^2 )/(a(√(a^2 +8bc))+b(√(b^2 +8ca))+c(√(c^2 +8ab)))) On the other hands,also by C−S we have a(√(a^2 +8bc))+b(√(b^2 +8ca))+c(√(c^2 +8ab))=(√a).(√(a^3 +8abc))+(√b).(√(b^3 +8abc))+(√c).(√(c^3 +8abc)) ≤(√((a+b+c)(a^3 +b^3 +c^3 +24abc))).Hence, P≥(((a+b+c)^2 )/(√((a+b+c)(a^3 +b^3 +c^3 +24abc))))=(√(((a+b+c)^3 )/(a^3 +b^3 +c^3 +24abc))).Therefore,it is enough to prove that (((a+b+c)^3 )/(a^3 +b^3 +c^3 +24abc))≥1⇔ (a+b+c)^3 ≥a^3 +b^3 +c^3 +24abc⇔(a+b)(b+c)(c+a)≥8abc This final inequality is always true because it is followed from the inequlities: a+b≥2(√(ab)) ,b+c≥2(√(bc)) ,c+a≥2(√(ca)) Thus,P≥1.The equality occurs if an only if a=b=c(q.e.d)

$$ \\ $$ $$ \\ $$ $$\boldsymbol{\mathrm{Applying}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{Cauchi}}−\boldsymbol{\mathrm{Schwartz}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{have}} \\ $$ $$\mathrm{we}\:\mathrm{have}\:\:\mathrm{P}=\frac{\mathrm{a}}{\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{8bc}}}+\frac{\boldsymbol{\mathrm{b}}}{\sqrt{\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{\mathrm{ca}}}}+\frac{\boldsymbol{\mathrm{c}}}{\sqrt{\boldsymbol{\mathrm{c}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{\mathrm{ab}}}}=\frac{\boldsymbol{\mathrm{a}}^{\mathrm{2}} }{\boldsymbol{\mathrm{a}}\sqrt{\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{\mathrm{bc}}}}+\frac{\boldsymbol{\mathrm{b}}^{\mathrm{2}} }{\boldsymbol{\mathrm{b}}\sqrt{\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{\mathrm{ca}}}}+\frac{\boldsymbol{\mathrm{c}}^{\mathrm{2}} }{\boldsymbol{\mathrm{c}}\sqrt{\boldsymbol{\mathrm{c}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{\mathrm{ab}}}}\geqslant\frac{\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)^{\mathrm{2}} }{\boldsymbol{\mathrm{a}}\sqrt{\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{\mathrm{bc}}}+\boldsymbol{\mathrm{b}}\sqrt{\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{\mathrm{ca}}}+\boldsymbol{\mathrm{c}}\sqrt{\boldsymbol{\mathrm{c}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{\mathrm{ab}}}}\: \\ $$ $$\boldsymbol{\mathrm{On}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{other}}\:\boldsymbol{\mathrm{hands}},\boldsymbol{\mathrm{also}}\:\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{C}}−\boldsymbol{\mathrm{S}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{have}} \\ $$ $$\boldsymbol{\mathrm{a}}\sqrt{\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{\mathrm{bc}}}+\boldsymbol{\mathrm{b}}\sqrt{\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{\mathrm{ca}}}+\boldsymbol{\mathrm{c}}\sqrt{\boldsymbol{\mathrm{c}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{\mathrm{ab}}}=\sqrt{\boldsymbol{\mathrm{a}}}.\sqrt{\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\mathrm{8}\boldsymbol{\mathrm{abc}}}+\sqrt{\boldsymbol{\mathrm{b}}}.\sqrt{\boldsymbol{\mathrm{b}}^{\mathrm{3}} +\mathrm{8}\boldsymbol{\mathrm{abc}}}+\sqrt{\boldsymbol{\mathrm{c}}}.\sqrt{\boldsymbol{\mathrm{c}}^{\mathrm{3}} +\mathrm{8}\boldsymbol{\mathrm{abc}}} \\ $$ $$\leqslant\sqrt{\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)\left(\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{b}}^{\mathrm{3}} +\boldsymbol{\mathrm{c}}^{\mathrm{3}} +\mathrm{24}\boldsymbol{\mathrm{abc}}\right)}.\boldsymbol{\mathrm{Hence}}, \\ $$ $$\boldsymbol{\mathrm{P}}\geqslant\frac{\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)^{\mathrm{2}} }{\sqrt{\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)\left(\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{b}}^{\mathrm{3}} +\boldsymbol{\mathrm{c}}^{\mathrm{3}} +\mathrm{24}\boldsymbol{\mathrm{abc}}\right)}}=\sqrt{\frac{\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)^{\mathrm{3}} }{\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{b}}^{\mathrm{3}} +\boldsymbol{\mathrm{c}}^{\mathrm{3}} +\mathrm{24}\boldsymbol{\mathrm{abc}}}}.\boldsymbol{\mathrm{Therefore}},\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{enough}} \\ $$ $$\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}\:\frac{\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)^{\mathrm{3}} }{\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{b}}^{\mathrm{3}} +\boldsymbol{\mathrm{c}}^{\mathrm{3}} +\mathrm{24}\boldsymbol{\mathrm{abc}}}\geqslant\mathrm{1}\Leftrightarrow \\ $$ $$\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)^{\mathrm{3}} \geqslant\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{b}}^{\mathrm{3}} +\boldsymbol{\mathrm{c}}^{\mathrm{3}} +\mathrm{24}\boldsymbol{\mathrm{abc}}\Leftrightarrow\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)\left(\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)\left(\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{a}}\right)\geqslant\mathrm{8}\boldsymbol{\mathrm{abc}} \\ $$ $$\boldsymbol{\mathrm{This}}\:\boldsymbol{\mathrm{final}}\:\boldsymbol{\mathrm{inequality}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{always}}\:\boldsymbol{\mathrm{true}}\:\boldsymbol{\mathrm{because}} \\ $$ $$\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{followed}}\:\boldsymbol{\mathrm{from}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{inequlities}}: \\ $$ $$\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\geqslant\mathrm{2}\sqrt{\boldsymbol{\mathrm{ab}}}\:,\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\geqslant\mathrm{2}\sqrt{\boldsymbol{\mathrm{bc}}}\:,\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{a}}\geqslant\mathrm{2}\sqrt{\boldsymbol{\mathrm{ca}}} \\ $$ $$\boldsymbol{\mathrm{Thus}},\boldsymbol{\mathrm{P}}\geqslant\mathrm{1}.\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{equality}}\:\boldsymbol{\mathrm{occurs}}\:\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{only}}\:\boldsymbol{\mathrm{if}} \\ $$ $$\boldsymbol{\mathrm{a}}=\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{c}}\left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right) \\ $$

Commented byFarruxjano last updated on 15/Jun/20

Sir thanks a lot

$$\mathrm{Sir}\:\mathrm{thanks}\:\mathrm{a}\:\mathrm{lot} \\ $$

Commented by1549442205 last updated on 25/Jun/20

Thank you!you are wellcome ,sir.

$$\mathrm{Thank}\:\mathrm{you}!\mathrm{you}\:\mathrm{are}\:\mathrm{wellcome}\:,\mathrm{sir}. \\ $$

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