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Question Number 78340 by loveineq. last updated on 16/Jan/20
Let a,b,c > 0 and c^2 = ((ab+bc+ca)/3) . Prove that ((a^3 +b^3 −2c^3 )/(a^3 +b^3 +c^3 )) ≤ 3(((a^2 +b^2 −2c^2 )/(a^2 +b^2 +c^2 )))
$$\mathrm{Let}\:\:{a},{b},{c}\:>\:\mathrm{0}\:\:\mathrm{and}\:\:{c}^{\mathrm{2}} \:=\:\frac{{ab}+{bc}+{ca}}{\mathrm{3}}\:.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} −\mathrm{2}{c}^{\mathrm{3}} }{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }\:\leqslant\:\mathrm{3}\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\right) \\ $$
Answered by MJS last updated on 16/Jan/20
(a^3 +b^3 −2c^3 )(a^2 +b^2 +c^2 )≤3(a^2 +b^2 −2c^2 )(a^3 +b^3 +c^3 ) ⇒ 2(a^5 +b^5 )−7c^2 (a^3 +b^3 )+5c^3 (a^2 +b^2 )+2a^2 b^2 (a+b)−4c^5 ≥0 let a=pc∧b=qc (2(p^5 +q^5 )+2p^2 q^2 (p+q)−7(p^3 +q^3 )+5(p^2 +q^2 )−4)c^5 ≥0 2(p^5 +q^5 )+2p^2 q^2 (p+q)−7(p^3 +q^3 )+5(p^2 +q^2 )−4≥0 3c^2 =ab+ac+bc 3c^2 =(p+pq+q)c^2 ⇒ q=((3−p)/(1+p)) ((2p^(10) +10p^9 +15p^8 −18p^7 −33p^6 −12p^5 −33p^4 −90p^3 +735p^2 −914p+338)/((p+1)^5 ))≥0 (p−1)^2 (2p^8 +14p^7 +41p^6 +50p^5 +26p^4 −10p^3 −79p^2 −238p+338)≥0 no other real linear factors with p=1 lhs=0 with p<>1 lhs>0 ⇒ proven
$$\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} −\mathrm{2}{c}^{\mathrm{3}} \right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\leqslant\mathrm{3}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{c}^{\mathrm{2}} \right)\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right) \\ $$$$\Rightarrow \\ $$$$\mathrm{2}\left({a}^{\mathrm{5}} +{b}^{\mathrm{5}} \right)−\mathrm{7}{c}^{\mathrm{2}} \left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)+\mathrm{5}{c}^{\mathrm{3}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({a}+{b}\right)−\mathrm{4}{c}^{\mathrm{5}} \geqslant\mathrm{0} \\ $$$$\mathrm{let}\:{a}={pc}\wedge{b}={qc} \\ $$$$\left(\mathrm{2}\left({p}^{\mathrm{5}} +{q}^{\mathrm{5}} \right)+\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{2}} \left({p}+{q}\right)−\mathrm{7}\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right)+\mathrm{5}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)−\mathrm{4}\right){c}^{\mathrm{5}} \geqslant\mathrm{0} \\ $$$$\mathrm{2}\left({p}^{\mathrm{5}} +{q}^{\mathrm{5}} \right)+\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{2}} \left({p}+{q}\right)−\mathrm{7}\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right)+\mathrm{5}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)−\mathrm{4}\geqslant\mathrm{0} \\ $$$$ \\ $$$$\mathrm{3}{c}^{\mathrm{2}} ={ab}+{ac}+{bc} \\ $$$$\mathrm{3}{c}^{\mathrm{2}} =\left({p}+{pq}+{q}\right){c}^{\mathrm{2}} \:\Rightarrow\:{q}=\frac{\mathrm{3}−{p}}{\mathrm{1}+{p}} \\ $$$$ \\ $$$$\frac{\mathrm{2}{p}^{\mathrm{10}} +\mathrm{10}{p}^{\mathrm{9}} +\mathrm{15}{p}^{\mathrm{8}} −\mathrm{18}{p}^{\mathrm{7}} −\mathrm{33}{p}^{\mathrm{6}} −\mathrm{12}{p}^{\mathrm{5}} −\mathrm{33}{p}^{\mathrm{4}} −\mathrm{90}{p}^{\mathrm{3}} +\mathrm{735}{p}^{\mathrm{2}} −\mathrm{914}{p}+\mathrm{338}}{\left({p}+\mathrm{1}\right)^{\mathrm{5}} }\geqslant\mathrm{0} \\ $$$$\left({p}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{2}{p}^{\mathrm{8}} +\mathrm{14}{p}^{\mathrm{7}} +\mathrm{41}{p}^{\mathrm{6}} +\mathrm{50}{p}^{\mathrm{5}} +\mathrm{26}{p}^{\mathrm{4}} −\mathrm{10}{p}^{\mathrm{3}} −\mathrm{79}{p}^{\mathrm{2}} −\mathrm{238}{p}+\mathrm{338}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{no}\:\mathrm{other}\:\mathrm{real}\:\mathrm{linear}\:\mathrm{factors} \\ $$$$\mathrm{with}\:{p}=\mathrm{1}\:\mathrm{lhs}=\mathrm{0} \\ $$$$\mathrm{with}\:{p}<>\mathrm{1}\:\mathrm{lhs}>\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{proven} \\ $$
Answered by loveineq. last updated on 16/Jan/20
Thanks MJS
$$\mathrm{Thanks}\:\mathrm{MJS} \\ $$

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