How prove that $10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$ let $a,b,c\ge 0$, such that $a+b+c=1$, prove that
$$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$$
This problem is simple as 2005, china west competition problem
$$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\ge 1$$
see:(http://www.artofproblemsolving.com/Forum/viewtopic.php?p=362838&sid=00aa42b316d41e251e24e658594fcc51#p362838)
for 2005 china west problem we have two methods (at least)
solution 1：
note 
$$(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(a+c)$$
$$(a+b+c)^5=a^5+b^5+c^5+5(a+b)(b+c)(a+c)(a^2+b^2+c^2+ab+bc+ac)$$
then 
$$\Longleftrightarrow 10[1-3(a+b)(b+c)(a+c)]-9[1-5(a+b)(b+c)(a+c)(a^2+b^2+c^2+ab+bc+ac)]\ge 1$$
$$\Longleftrightarrow 3(a+b)(b+c)(a+c)(a^2+b^2+c^2+ab+bc+ac)-2(a+b)(b+c)(a+c)\ge 0$$
$$\Longleftrightarrow 3(a^2+b^2+c^2+ab+bc+ac)\ge 2=2(a+b+c)^2$$
$$ a^2+b^2+c^2-ab-bc-ac\ge 0$$
It's Obviously.
solution 2:
$$10(a+b+c)^2(a^3+b^3+c^3)-9(a^5+b^5+c^5)-(a+b+c)^5\ge0$$
it is equivalent to
$$15(a+b)(b+c)(c+a)(a^2+b^2+c^2-ab-bc-ca)\ge0$$
But for
 $$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$$
and  for this equality I think 
$$10a^3-9a^5\le p (a-1/3)+q$$
and let
$$f(x)=10x^3-9x^5\Longrightarrow f'(x)=30x^2-45x^4\Longrightarrow p=f'(1/3)=\dfrac{25}{9}$$
$$q=f(1/3)=\dfrac{1}{9}$$
so if we can prove this
$$10a^3-9a^5\le\dfrac{25}{9}(a-1/3)+\dfrac{1}{9}$$
These methods I can't work, can someone help deal it. Thank you 
 A: EDIT: The original proof contained an error. It is (hopefully) fixed now.

Consider the function $f(x) = 10x^3 - 9x^5$. Then our goal is to show that $f(a) + f(b) + f(c) \leq 9/4$.
We will need the following two claims
Claim1: $f(x) + f(1-x) \leq 9/4$ for all $x \in [0,1]$.
Proof: A straightforward calculation says that the local maximum is obtained in $x = 0.5 \pm \frac{1}{2\sqrt{3}}$ and is equal to $9/4$.
Claim2: For all $0 \leq a \leq b$ such that $a+b \leq 2/3$ we have $f(a)+f(b) \leq f(a+b)$ 
Proof: The claim is trivial if $a=0$. Therefore, we shall assume that $a>0$.We need to prove that
$$
10a^3 - 9a^5 + 10b^3 - 9b^5 \leq 10(a+b)^3 - 9(a+b)^5.
$$
Opening the parenthesis on the RHS, and reducing we get that the above is equivalent to
$$
0 \leq 10(3a^2b + 3ab^2) - 9(5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4)
$$
Since $a,b > 0$, we can divide by $15ab$, and so, by moving sides it is enough to show that
$$
3(a^3 + 2a^2b + 2ab^2 + b^3) \leq 2(a + b).
$$
Adding $3(a^2b+ab^2)$ to both sides we get
$$
3(a+b)^3 \leq (2+3ab)(a + b).
$$
Since $a,b \geq 0$, we can divide by $a+b$ to get
$$
3(a+b)^2 \leq 2+3ab,
$$
or equivalently
$$
3a^2+3b^2 + 3ab \leq 2
$$
It is easy to check that the inequality holds if $a+b \leq 2/3$.

We now turn to the proof. Let's s assume that $a \leq b \leq c$.
Since $a+b \leq 2/3$, by Claim2 we have $f(a)+f(b) \leq f(a+b)$, and therefore, by Claim1 we get $f(a)+f(b)+f(c) \leq f(a+b) + f(c) = f(1-c) + f(c) \leq 9/4$, as required.
A: Set $c=1-a-b$ and find the three lines on which $\frac{\partial f(a,b)}{\partial a\partial b}=0$. Now derive $f(a,b(a))$ according to $a$ to find the extrema. This is a bit tedious, but you fill get the desired $9/4$.
A: We need to prove that
$$40(a^3+b^3+c^3)(a+b+c)^2-36(a^5+b^5+c^5)\leq9(a+b+c)^5$$ and since the last inequality is homogeneous already,
it's enough to prove this inequality for all non-negatives $a$, $b$ and $c$.
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, since our inequality is fifth degree, we need to prove that $f(w^3)\geq0,$ where $f$ is a linear function.
But the linear function gets a minimal value for the extreme value of $w^3$, 
which happens in the following cases.


*

*$w^3=0$. 


Let $c=0$ and $b=1$.
We obtain:
$$(a^2-4a+1)^2\geq0;$$
2. Two variables are equal.
Let $b=c=1$.
We obtain:
$$9(a+2)^5\geq40(a^3+2)(a+2)^2-36(a^5+2)$$ or
$$a^5-14a^4+40a^3+128a^2+80a+8)\geq0,$$ which is true by AM-GM:
$$a^5-14a^4+40a^3+128a^2+80a+8\geq$$
$$\geq a^2\left(3\left(\frac{a^3}{3}\right)+40a+128-14a^2\right)\geq a^4\left(5\sqrt[5]{\left(\frac{1}{3}\right)^3\cdot40\cdot128}-14\right)\geq0.$$
Done!
