mateforum.ro

Daca \( a,b,c > 0 \) si \( abc=1 \), aratati ca

\( a^2(b^5+c^5) + b^2(a^5+c^5) + c^2(b^5+a^5) \geq 2(a+b+c) \)

Dragoi Marius, Posa Bogdan (revista Cardinal)

Folosind faptul ca abc=1, inegalitatea se rescrie \( (a^3+b^3+c^3)(\frac{1}{c^2}+\frac{1}{b^2}+\frac{1}{a^2})\geq3(a+b+c) \), adevarata in baza inegalitatii lui Cebasev.

\( \sum_{ciclic}^{}({a^2(b^5+c^5)}) \geq \sum_{ciclic}^{}{{\frac{({b+c})({b^4+c^4})}{2}}{a^2} \geq \sum_{cilcic}^{}({a^2b^2c^2(b+c)}) = \sum_{cilcic}^{}({b+c}) = 2(a+b+c) \).

Vrem ca: \( \sum_{sym}a^5b^2\geq 2(a+b+c)=\sum_{sym}a^3b^2c^2 \), care reiese din Inegalitatea lui Muirhead, deoarece
tripletul \( (5;2;0) \) majorizeaza tripletul \( (3;2;2). \) Cel mai probabil asa a fost dedusa inegalitatea.