This resembles APMO 2004/5 so much. Note that if $x>0$ then $x^5-x^2+3 \geq x^3+2$, which can be seen by factoring into $(x^3-1)(x^2-1) \geq 0$. Then the inequality reduces to \[ \prod (a^3 +2 ) \geq (a+b+c)^3 \] or, \[ \prod (\frac{a^3-1}{3}+1) \geq (\frac{a+b+c}{3})^3 \] If $a,b,c \geq 1$, or $a,b,c \leq 1$, then by Bernoulli's Inequality, we have \[ \prod (\frac{a^3-1}{3}+1) \geq 1 + \sum \frac{a^3-1}{3} =\sum a^3/3 \] Complete by Power-Mean If $a \geq 1 \geq b,c$ or $a \leq 1 \leq b,c$, then by Bernoulli and Holder, we have \[ \prod (\frac{a^3-1}{3}+1) \geq (\frac{a^3+1+1}{3})(1+\frac{b^3-1}{3}+\frac{c^3-1}{3})\geq (\frac{a^3+1+1}{3})(\frac{1+b^3+c^3}{3})(\frac{1+1+1}{3}) \geq (\frac{a+b+c}{3})^3 \] Q.E.D. Note: Bernoulli's inequality (or perhaps an extension of it, anyone know its name?) states that if $a_1, \ldots, a_n > -1$, and that all the {a} have the same sign, then $\prod(1+a) \geq 1+\sum a$.

After Yufei's brilliant observation that x^5 - x^2 + 3 >= x^3 + 2, the proof results more directly by application of the generalized Holder Inequality to the matrix a^3 1 1 1 b^3 1 1 1 c^3. Assigning weights of 1/3 to each row gives [(a^3+1+1)(1+b^3+1)(1+1+c^3)]^1/3>=(a+b+c)^3 as required.

Can you please send all USAMO 2004 problems? I would be very grateful to you.

So I had to use the hint from the AoPS book, which was as follows. After that, the problem pretty much follows.

The following inequalities are also true. Let $a,b,c>0$ , prove that\[(a^5-a^3+3)(b^5-b^3+3)(c^5-c^3+3)\geq 3(a+b+c)^2.\] here Let $a ,b, c$ be positive real numbers , prove that\[ \left(a^{2013}-a^{2011 }+ 3\right)\left(b^{2013}-b^{2011}+ 3\right)\left(c^{2013}-c^{2011} + 3\right)\geq 9\left(a^2 + b^2 + c^2\right) .\] here

In case anyone wants some motivation as to where the inequality $a^5 - a^2 + 3 \geq a^3 + 2$, here is how we might think of something like this. Note that the RHS of the inequality is of degree 3, so we want to prove some inequality of the form $a^5 - a^2 + 3 \geq ka^3 + \ell$, for some real numbers $k$ and $\ell$. Replace $a^3$ with $x$, so we want to prove an inequality of the form $x^\frac{5}{3} - x^\frac{2}{3} + 3 \geq kx + \ell$. Note that we have equality in the original inequality at $a = 1$, so we want equality here when $x = 1$. This suggest making the line $y = kx + \ell$ tangent to this curve at the point $(1,3)$. the derivative of $x^\frac{5}{3} - x^\frac{2}{3} + 3$ evaluated at $1$ turns out just to be $1$, so the equation of the tangent line at $(1,3)$ is $y - 3 = x - 1$, or $y = x + 2$, so putting back in $x = a^3$, we want to prove that $a^5 - a^2 + 3 \geq a^3 + 2$, where is very easy to do, just by factoring, and then the final inequality we get follows easily from holders inequality.

never mind, incorrect solution was here thanks mssmath

The Jensen solution above is definitely wrong as the sign of Jensen's on f(3x) is reversed.

Solutions of problem in this pack.

TheStrangeCharm wrote: In case anyone wants some motivation as to where the inequality $a^5 - a^2 + 3 \geq a^3 + 2$, here is how we might think of something like this. Note that the RHS of the inequality is of degree 3, so we want to prove some inequality of the form $a^5 - a^2 + 3 \geq ka^3 + \ell$, for some real numbers $k$ and $\ell$. I don't understand how you got this. How did you get rid of the non $a^3$ terms?

Is the observation related to the quantity in question being greater than or equal to $a^3+3$ necessary? If so, how did people manage to discover this while solving the problem?

Delray wrote: Is the observation related to the quantity in question being greater than or equal to $a^3+3$ necessary? If so, how did people manage to discover this while solving the problem? This is motivated by holders (three var suggests holder)

Binomial-theorem wrote: So I had to use the hint from the AoPS book, which was as follows. After that, the problem pretty much follows.

does anybody know the motivation for the hint?

bump plz help

@above We wanted to remove the power 5 term along with the squares so that even if some cubics are left we can simly use holder on it so $(a^2-1)(a^3-1)$ was the best candidate.

Notice that $a^5 - a^3 - a^2 + 1 = a^3(a^2-1) - (a^2-1) = (a^3-1)(a^2-1) = (a+1)(a-1)^2(a^2+a+1) \geq 0$ since a is positive. $a^5 - a^2 + 1 \geq a^3$ $a^5 - a^2 + 3 \geq a^3 + 2$ $(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \geq (a^3 + 2)(b^3+2)(c^3+2)$ By Holder's inequality, $\sqrt{(a^3+1+1)(b^3+1+1)(c^3+1+1)} \geq (a+b+c)$ $(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \geq (a^3 + 2)(b^3+2)(c^3+2) \geq (a+b+c)^2$ and we are done.

Observe that $(a^2-1)(a^3-1) \ge 0$, so $a^5-a^2-a^3+1 \ge 0$. This can be written as $a^5-a^2+3 \ge a^3+2$. Hence, it suffices to prove \[\prod (a^3+2) \ge (a+b+c)^3.\]However, \[\prod (a^3+1+1)^{\frac{1}{3}} \ge (a+b+c)\]by Holder's, so \[\prod(a^3+2) \ge (a+b+c)^3\]by cubing both sides and we're done.

I think the step to find $a^5-a^2+3 \geq a^3+2$ is pretty natural? It's pretty obvious that the fifth powers don't relate to the problem at all, so I decided to tame it with third powers so possibly I could use an inequality to finish. But alas, I didn't find the nice solution. Bad Solution. Observe that \begin{align*} (a^3-1)(a^2-1) \geq 0 \ \forall a > 0 \\ a^5-a^2+3 \geq a^3+2 \ \forall a > 0 \end{align*}Thus it suffices to show that $\prod_{cyc} (a^3+2) \geq (a+b+c)^3$. Expanding, the inequality is equivalent to $$a^3b^3c^3+2\sum_{cyc} a^3b^3+3\sum_{cyc} a^3 - 3\sum_{sym} a^2b-6abc+8 \geq 0.$$By Schur, $$3\sum_{cyc} a^3+9abc-3\sum_{sym} a^2b \geq 0,$$and by AM-GM, $$2\sum_{cyc} a^3b^3 \geq 6a^2b^2c^2.$$Thus we need $$a^3b^3c^3+6a^2b^2c^2-15abc+8\geq 0 \iff (abc-1)^2(abc+8) \geq 0,$$which is clearly true for $abc > 0$. Equality holds when $a=b=c=1$. $\square$

Claim: $x^3+2 \le x^5-x^2+3$ Proof: We have $$0 \le (x^3-1)(x^2-1)= x^5-x^3-x^2+1 \implies x^3+2 \le x^5-x^2+3.$$$\Box$ Claim: $\sum_{cyc} (a^3+2 ) \le (a+b+c)^3.$ Proof: Expand the LHS and RHS, so the desired inequality is the same as $$a^3b^3c^3+3(a^3+b^3+c^3)+2(a^3b^3+b^3c^3+c^3a^3)+8 \ge 3(a^2b+b^2a+b^c+c^2b+a^2c+c^2a)+6abc.$$Now we use the fact that $$a^3+a^3b^3+1 \ge 3a^2b,$$this is direct from AM-GM. After using this, our original inequality can be simplified down to $$a^3b^3c^3+a^3+b^3+c^3+1+1 \ge 6abc,$$but this is also straightforward from AM-GM, therefore, our original inequality is true and we are done. $\blacksquare$

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Note that $(x^3-1)(x^2-1) \ge 0 \implies x^5-x^2+3 \ge x^3+2$, by expansion and rearranging. Thus $(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3) \ge (a^3+1+1)(1+b^3+1)(1+1+c^3) \ge (a+b+c)^3$, by Holder's.

We will prove a stronger result. Note that $(x^3 - 1)(x^2 - 1) \ge 0$ implies that $x^5 - x^2 \ge x^3 - 1$. So we need to prove \[(a^3 +2)(b^3 + 2)(c^3 + 2) \ge (a+b+c)^3.\]By Holder's we have \[(a^5 - a^2 + 3)(b^5 - b^2 +3)(c^5 - c^2 + 3) \ge (a^3 + 1 + 1)(1+b^3+1)(1+1+c^3) \ge (a+b+c)^3 \blacksquare\]

Pretty much same as mathpi's solution oops. Theres probably a cleaner finish using Holder's. It is trivial to see that $(x^3-1)(x^2-1) \geq 0 \implies x^5-x^2+3 \geq x^3+2$. It suffices to prove $\prod_{\text{cyc}} a^3+2 \geq (a+b+c)^3$. Expanding, and simplifying, we receive $-a^3b^3c^3-2a^3b^3-2a^3c^3-3a^3+3a^2b+3a^2c+3ab^2+6abc+3ac^2-2b^3c^3-3b^3+3b^2 c+3bc^2-3 c^3-8 \leq 0$. Now notice that $$a^3+a^3b^3+1 \geq 3a^2b$$by AM-GM, which simplifies the inequality to $a^3b^3c^3+a^3+b^3+c^3+1+1 \geq 6abc$ which is a direct result of AM-GM You could also use Holder's as the last two posts stated.

Notice $(a-1)^2(a+1)(a^2+a+1)\ge 0$ so $a^5-a^2+3\ge a^3+2.$ Hence, $$\prod_{\text{cyc}}(a^5-a^2+3)\ge\prod_{\text{cyc}}(a^3+1+1)\ge (a+b+c)^3$$by Holder. $\square$

Let $a, b, c > 0$. Prove that $$(a^2-a+3)(b^2-b+3)(c^2-c+3)\geq 9(a+b+c) $$$$(3a^2-2a+1)(3b^2-2b+1)(3c^2-2c+1)\geq \frac{3}{16}(a+b+c)^2$$$$(5a^2-2a+3)(5b^2-2b+3)(5c^2-2c+3)\geq \frac{72}{5}(a+b+c)^2$$

Arne wrote: Let $a, b, c > 0$. Prove that $(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \geq (a + b + c)^3$.

sqing wrote: Arne wrote: Let $a, b, c > 0$. Prove that $(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \geq (a + b + c)^3$. Typos ($(a^5+4)(b^5+4)(c^5+4) \ge (a + b + c + 2)^3$ is not true). I think it should be $$(3/5)^3(a^5 + 4)(b^5 + 4)(c^5 + 4)\ge (3/5)^3 (a+b+c+2)^5/5^2 \ge \cdots$$

we can first simplify $a^5-a^2+3$ as $a^5-a^2-a^3+1+2+a^3=(a^3-1)(a^2-1)+2+a^3$, and we can easily prove that $(a^3-1)(a^2-1) \geq 0$ and $a^5-a^2+3 \geq a^3+2$ doing the same for $b$ and $c$, and then plugging it in, we get that $(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \geq (a^3+2)(b^3+2)(c^3+2)$ we can easily see that $\sqrt[3]{(a^3+1+1)(b^3+1+1)(c^3+1+1)} \geq (a+b+c)$, and $(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \geq (a^3+1+1)(b^3+1+1)(c^3+1+1) \geq (a+b+c)^3$ from hölder's inequality

By a fuction ......

The original problem is a special case of the following generalization where $$p=2$$.

Generalization 1 Let $a_{1},a_{2},\cdots,a_{p+1}$ be positive reals and $p$ be a nonnegative integer. Then prove that $$\prod{\left(a^{2p+1}-a^{p}+p+1\right)}\geq \left(\sum_{cyc}{a_{1}}\right)^{p+1}$$

Note that $x^5-x^2+3\ge x^3+2$ since this is equivalent to $(x-1)^2(x^2+x+1)(x+1)\ge 0$. So, it suffices to prove \[(a^3+2)(b^3+2)(c^3+2)\ge (a+b+c)^3\]and by Holder's, we have \[(a^3+1+1)(1+b^3+1)(1+1+c^3)\ge (a+b+c)^3\]so we are done. $\square$

The main claim is that \[a^5 - a^2 + 3 \geq a^3 + 2\]which is true as it rearranges to \[(a - 1)^2 (a + 1)(a^2 + a + 1) \geq 0\text{.}\]Now we have \begin{align*} (a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) &\geq (a^3 + 2)(b^3 + 2)(c^3 + 2) \\ &= (a^3 + 1 + 1)(1 + b^3 + 1)(1 + 1 + c^3) \\ &\geq (a + b + c)^3 \end{align*}by Hölder's inequality.

Notice that \[a^5-a^2+3 \ge a^3+1+1 \iff (a^2-1)(a^3-1) \ge 0\] holds over the positive reals. We finish using Holder's, which tells us \[(a^3+1+1)(1+b^3+1)(1+1+c^3) \ge (a+b+c)^3. \quad \blacksquare\]

I have discussed this problem on my Holder video in the inequality tutorial playlist on my channel "Little Fermat". Here is the Video