We need to prove : $10(a^3+b^3+c^3)(a+b+c)^2-9(a^5+b^5+c^5) \geq (a+b+c)^5$ $\iff 15\sum_{sym}(a^4b-a^2b^2c) \geq 0$ : obviously true by Muirhead

That is an obvious way, but what about other ways?

An elegant factorization here.

Can we prove that $abc\geq ab+bc+ca\ or\ not?$ If it is possible, then I can show you another solution. kunny

No, it's false.That one would be equivalent whith 1>=1/a+1/b+1/c.But because a+b+c=1, we have, for example, a<1.Then 1/a>1.=>1/a+1/b+1/c>1, contradiction.

Let $a\ge b\ge c$, so $\frac{1}{2}\ge b\ge c$, $10(a^{3}+b^{3}+c^{3}) - 9(a^{5}+ b^{5} + c^{5})-1$ $=\frac{15}{16}(b+c)(b-c)^2[8-3(3b^2+2bc+3c^2)]+\frac{15}{16}(1-a)(3a-1)^2(a+1)^2\ge 0$ $\Rightarrow10(a^{3}+b^{3}+c^{3}) - 9(a^{5}+ b^{5} + c^{5})\ge1$ http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=518357&p=2920282#p2920282

Skyward_Sea wrote: If $a,b,c$ are positive reals such that $a+b+c=1$, prove that \[ 10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\geq 1 . \] We have: $\Leftrightarrow 10\left( {{{\left( {3a} \right)}^3} + {{\left( {3b} \right)}^3} + {{\left( {3c} \right)}^3}} \right) - \left( {{{\left( {3a} \right)}^5} + {{\left( {3b} \right)}^5} + {{\left( {3c} \right)}^5}} \right) \ge 27$ Let $3a=x, 3b=y, 3c=z \Rightarrow x,y,z > 0\,\,\,,\,x + y + z = 3$ Case 1. $x,y,z \in \left( {0;2} \right]$ and note that $10{t^3} - {t^5} \ge 25t - 16,\,\,\forall t \in \left( {0;2} \right]$ Case 2. Exist $x>2$, let $f(t)=10t^3-t^5, t \in \left( {2;3} \right)$ $f'\left( t \right) = 30{t^2} - 5{t^4},\,\,f'\left( t \right) = 0 \Leftrightarrow t = \pm \sqrt {\frac{6}{5}} $ $f\left( t \right) > f\left( 3 \right) = 27$ in other hand $10y^3>y^5, 10z^3>z^5 $ $ \Rightarrow 10{x^3} - {x^5} + 10{y^3} - {y^5} + 10{z^3} - {z^5} > 27$.

Skyward_Sea wrote: If $a,b,c$ are positive reals such that $a+b+c=1$, prove that \[ 10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\geq 1 . \] Suppose $c = \max\{a,b,c\},$ set $t = \frac{a+b}{2}$ and \[f(a,b,c) = 10(a^3+b^3+c^3)-9(a^5+b^5+c^5).\]We have \[f(a,b,c) - f(t,t,c) = \frac{15}{16}(a+b)[8 - 3(a+b)^2 -6(a^2+b^2)](a-b)^2.\]Because $a + b \leqslant \frac{2}{3}$ so $3(a+b)^2 + 6(a^2+b^2) \leqslant 8,$ therefore $f(a,b,c) \geqslant f(t,t,c).$ Another, we have \[f(t,t,c) = f(t,t,1-2t) = 1 + 30t(3t-1)^2(t-1)^2 \geqslant 1.\]The proof is completed.

Skyward_Sea wrote: If $a,b,c$ are positive reals such that $a+b+c=1$, prove that \[ 10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\geq 1 . \] By computer，we have $$LHS-RHS=\frac{5}{2}\sum_{cyc}{(b+c)(bc+a)(3a-1)^2}\ge{0}$$

We need to prove $10(\sum x^3 - (\sum x)^3) \geq 9 (\sum x^5 - (\sum x)^5)$ Which is equivalent to $-30\prod(x+y) \geq -45\prod(x+y)(\sum x^2+\sum xy)$ $\leftrightarrow x^2+y^2+z^2+xy+yz+zx \geq \frac{2}{3}$ $\leftrightarrow 1 - \sum xy \geq \frac{2}{3} \leftrightarrow \sum xy \leq \frac{1}{3}$ which is true because $1 = (x+y+z)^2 \geq 3\sum xy$

Just expand all And get.. $a^2+b^2+c^2\ge (ab+bc+ac)$ Which is true