How prove that $10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$

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 compition 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(or most two)
solution 1F 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 $$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$$
This methods I can't work, can someone help deal it.Thank you