6

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

Michael Rozenberg
  • 203,855
math110
  • 94,932
  • 17
  • 148
  • 519
  • Using symmetric polynomials $q = ab + bc + ca$ and $r = abc$, the inequality is $\displaystyle (q-r)(1-3q) \le \frac{1}{12}$. Will try later to show this if someone doesn't get it. – Macavity Aug 18 '13 at 11:24
  • Unfortunately, the inequality $10a^3-9a^5\le\dfrac{25}{9}(a-1/3)+\dfrac{1}{9}$ fails for all $0\le a\le 9/10$, see the graph – Alex Ravsky Aug 16 '17 at 13:07

3 Answers3

5

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.

Igor Shinkar
  • 841
  • Your case 2 needs a bit more work. Technically, the sufficient condition that you need is for $b+c$ to lie in the convex range of the function, so does not hold for $b+c \geq 0.578$. I.e., it does not follow from Claim 2 that $f(0.4) + f(0.2) \leq f(0.6)$. It could likely be changed to $f(b) + f(c) \leq f(0.5) + f(b+c-0.5) \leq f(b+c)$. – Calvin Lin Aug 18 '13 at 16:24
  • Thank you, I have consider a very very nice methods. – math110 Aug 19 '13 at 01:40
  • $f(b)+f(c) \le f(b+c)$ has nothing to do with convexity which instead implies $f(b)+f(c) \ge 2f((b+c)/2)$. – Phira Aug 19 '13 at 08:57
  • @CalvinLin and Phira, you are right. I fixed the solution – Igor Shinkar Aug 21 '13 at 07:14
  • 1
    @Phira Convexity does imply that $f(b) + f(c) \leq f(b+c) + f(0) = f(b+c)$. Apart from knowing the minimum, we also know the maximum value. – Calvin Lin Aug 21 '13 at 14:47
3

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 $9/4$.

Nathaniel Bubis
  • 33,425
0

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.

  1. $w^3=0$.

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!

Michael Rozenberg
  • 203,855