6
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Macavity Aug 18 '13 at 11:24
  • $\begingroup$ 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 $\endgroup$ – Alex Ravsky Aug 16 '17 at 13:07
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $f(b) + f(c) \leq f(0.5) + f(b+c-0.5) \leq f(b+c)$. $\endgroup$ – Calvin Lin Aug 18 '13 at 16:24
  • $\begingroup$ Thank you, I have consider a very very nice methods. $\endgroup$ – math110 Aug 19 '13 at 1:40
  • $\begingroup$ $f(b)+f(c) \le f(b+c)$ has nothing to do with convexity which instead implies $f(b)+f(c) \ge 2f((b+c)/2)$. $\endgroup$ – Phira Aug 19 '13 at 8:57
  • $\begingroup$ @CalvinLin and Phira, you are right. I fixed the solution $\endgroup$ – Igor Shinkar Aug 21 '13 at 7:14
  • 1
    $\begingroup$ @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. $\endgroup$ – Calvin Lin Aug 21 '13 at 14:47
3
$\begingroup$

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 $9/4$.

$\endgroup$
  • $\begingroup$ Thank you ,But I dislike this Lag $\endgroup$ – math110 Aug 18 '13 at 8:50
  • $\begingroup$ Why the downvote? $\endgroup$ – nbubis Aug 18 '13 at 8:57
  • $\begingroup$ Sorry, NO, I only see this answer is not true usefull(or not nice ), then I can downvoted it, $\endgroup$ – math110 Aug 18 '13 at 9:13
0
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.