parmenides51 wrote: Prove that $(a+b+c)^5 \ge 81 (a^2+b^2+c^2)abc$ for any positive real numbers $a,b,c$ I.Gorodnin It's a very old Vasc's inequality.

parmenides51 wrote: Prove that $$(a+b+c)^5 \ge 81 (a^2+b^2+c^2)abc$$for any positive real numbers $a,b,c$ I.Gorodnin Let $a,b,c $ be positive real numbers . Prove that$$( a+b) ( b+c)( c+a)( a+b+c) ^{2}\ge 24(a^2+b^2+c^2)abc$$ https://artofproblemsolving.com/community/c6h1282022p6753168 https://artofproblemsolving.com/community/c6h514188p2888532 https://artofproblemsolving.com/community/c6h87919p522653

parmenides51 wrote: Prove that $(a+b+c)^5 \ge 81 (a^2+b^2+c^2)abc$ for any positive real numbers $a,b,c$ I.Gorodnin Let $Q={\frac {27}{8}}\, \left( b+c \right) \left( c+a \right) \left( a+b\right) \left( a+b+c \right) ^{2}$ Then we have $$(a+b+c)^5 \geq Q \geq 81 (a^2+b^2+c^2)abc$$ $(a+b+c)^5 \geq Q$ use AG-GM proved. and $$Q\geq 81 (a^2+b^2+c^2)abc$$<=> $${\frac {27}{8}}\, \left( b-c \right) ^{2} \left( 3\,a-b-c \right) ^{2} a+{\frac {27}{8}}\, \left( c-a \right) ^{2} \left( 3\,b-c-a \right) ^{ 2}b+{\frac {27}{8}}\, \left( a-b \right) ^{2} \left( 3\,c-a-b \right) ^{2}c\geq 0 $$Done!

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@below this is wrong btw. $p^3-81r$ can be negative so using $p^2\geq 3q$ doesn't work for all actually.

parmenides51 wrote: Prove that $(a+b+c)^5 \ge 81 (a^2+b^2+c^2)abc$ for any positive real numbers $a,b,c$ I.Gorodnin A bit stronger version \[(a+b+c)^5 \geqslant (a^2+b^2+c^2)(a^3+b^3+c^3+78abc).\]

Very nice