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

Question

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

Answer 1

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 $9/4$.

Answer 2

EDIT: The original proof contained an error. It is (hopefully) fixed now.

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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 $a+b \leq 2/3$.

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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.