Let $p=\sum a^2, q=\sum ab$. This is: \[2(p^{2012}-q^{2012})\le \sum (p^{2012}-(a^2+2bc)^{2012})\] \[\Leftrightarrow (\sum (a-b)^2)[p^{2011}+p^{2010}q+...+q^{2011}]\le \\ \sum (b-c)^2[p^{2011}+p^{2010}(a^2+2bc)+...+(a^2+2bc)^{2012}]\] Let $s_a=a^2+2bc$ from now on: \[\Leftrightarrow \sum (b-c)^2[p^{2010}[s_a-q]+\\p^{2009}[s_a^2-q^2]+...+[s_a^{2011}-q^{2011}]]\ge 0\] \[\Leftrightarrow \sum (a-b)(a-c)(b-c)^2[p^{2010}+p^{2009}[s_a+q]+...\\+[s_a^{2010}+s_a^{2009}q+...+q^{2010}]]\ge 0\] Note that factoring out the $p'$s and $q'$s as necessary, it suffices to show that: \[\sum (a-b)(a-c)(b-c)^2s_a^k\ge 0\] For all nonnegative integers $k$. WLOG let $a\ge b\ge c$, and set $a=b+x, c=b-y$. Then this is: \[ys_a^k+xs_c^k\ge (x+y)s_b^k\] \[\Leftrightarrow y(s_a^k-s_b^k)\ge x(s_b^k-s_c^k)\] Assume $k>0$, since otherwise this is obvious. Then it's: \[\Leftrightarrow y(a-b)(a+b-2c)[s_a^{k-1}+s_a^{k-2}s_b+...+s_b^{k-1}]\ge \\x(b-c)(a+c-2b)[s_b^{k-1}+s_b^{k-2}s_c+...+s_c^{k-1}]\] \[\Leftrightarrow (x+y)[s_a^{k-1}+s_a^{k-2}s_b+...+s_b^{k-1}]\ge \\ (x-y)[s_b^{k-1}+s_b^{k-2}s_c+...+s_c^{k-1}]\] If $x<y$, this is trivially true. Otherwise it follows from $x+y\ge x-y$ and $s_a\ge s_c$. $\blacksquare$ Equality holds if $a=b$ or $b=c$ or $c=a$.

applepi2000 wrote:

And that is why this problem wasn't on the ELMO. (Among other things.)

How about:

meowme wrote: majorizes $[a^2 + 2bc, b^2 + 2ac, c^2 + 2ab]$. This isn't sorted correctly... Take $a=9001$, $b=1$ and $c=0$ for instance. applepi2000 wrote:

This inequality can be proved using induction and global derivative in few lines, but that makes problem not suitable for olympiad. I'd like to know what the official solution is. Proof: We will prove inequality for all $a,b,c\ge 0$ and $k \in \mathbb{N}$ $\sum (a^2+2bc)^k \leq (a^2+b^2+c^2)^k+2(ab+ac+bc)^k \quad (1)$. If $k=1$, we have equality. Now suppose it holds for some $k-1$. Now, we define $F(a,b,c)=RHS-LHS$ in $(1)$. If we denote global deribvative with $[F]$, then: $[F]=2k(a+b+c)\left[(a^2+b^2+c^2)^{k-1}+2(ab+ac+bc)^{k-1}-\sum (a^2+2bc)^{k-1} \right]$ Because of assumption, we have $[F]\ge 0$. Now, using global derivative theorem, it is enough to prove inequality when $abc=0$. WLOG we can suppose $c=0$, so it remains to prove: $a^{2k}+b^{2k}+(2ab)^{k}\le (a^2+b^2)^k+2(ab)^k$ $\Leftrightarrow (2^k-2)(ab)^k \le \sum_{i=1}^{k-1} \dbinom{k}{i}a^{2i}b^{2(k-i)}$ What follows from Am-Gm. Equality holds when $a=b=c$ and when $a=b, \; c=0$ with all permutations. $\blacksquare$

See also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2749340#p2749340

math154 wrote: Let $a,b,c\ge0$. Show that $(a^2+2bc)^{2012}+(b^2+2ca)^{2012}+(c^2+2ab)^{2012}\le (a^2+b^2+c^2)^{2012}+2(ab+bc+ca)^{2012}$. Calvin Deng. Here is a proof by AM-GM inequality. We will prove the general inequality If $a,b,c$ are positive real numbers and $n$ is a positive integer, then \[ (a^2+2bc)^{n}+(b^2+2ca)^{n}+(c^2+2ab)^{n} \leq (a^2+b^2+c^2)^{n}+2(ab+bc+ca)^{n}.\] For $n=1,$ and $n=2,$ the inequality become the equality. Consider the case $ n\geq 3.$ Setting $ a^2+2bc=A,\ b^2+2ca=B,\ c^2+2ab=C,\ a^2+b^2+c^2=D,\ ab+bc+ca=E,$ the inequality becomes \[ A^n+B^n+C^n \leq D^n+2E^n,\] \[ (D^n-A^n)+(D^n-B^n)+(D^n-C^n) \geq 2(D^n-E^n),\] \[{ \sum \left[(b-c)^2\left(D^{n-1}+D^{n-2}A+...+DA^{n-2}+A^{n-1}\right)\right] \geq \left[\sum(b-c)^2\right]\left(D^{n-1}+D^{n-2}E+...+DE^{n-2}+E^{n-1}\right).}\] This inequality follows from the following inequality \[ D^{n-k-1}\left[(b-c)^2(A^k-E^k)+(c-a)^2(B^k-E^k)+(a-b)^2(C^k-E^k)\right] \geq 0,\] where $ k=1,\ ... \ ,\ n-1.$ Using the AM-GM inequality, we have \[ A^k+(k-1)E^K \geq kAE^{k-1} \Longrightarrow A^k-E^k \geq kE^{k-1}(A-E)=kE^{k-1}(a-b)(a-c).\] Therefore \[ { \sum (b-c)^2(A^k-E^k) \geq kE^{k-1}\left[(b-c)^2(a-b)(a-c)+(c-a)^2(b-c)(b-a)+(a-b)^2(c-a)(c-b)\right]=0.}\] It completed the proof. Equality holds if and only if $(a-b)(b-c)(c-a)=0.\ \blacksquare$

applepi2000 wrote: Then it's: $\Leftrightarrow y(a-b)(a+b-2c)[s_a^{k-1}+s_a^{k-2}s_b+...+s_b^{k-1}]\ge \\x(b-c)(a+c-2b)[s_b^{k-1}+s_b^{k-2}s_c+...+s_c^{k-1}]$ $\Leftrightarrow (x+y)[s_a^{k-1}+s_a^{k-2}s_b+...+s_b^{k-1}]\ge \\ (x-y)[s_b^{k-1}+s_b^{k-2}s_c+...+s_c^{k-1}]$ If $x<y$, this is trivially true. Otherwise it follows from $x+y\ge x-y$ and $s_a\ge s_c$. $\blacksquare$ Equality holds if $a=b$ or $b=c$ or $c=a$. Your proof is not correct but the thing you prove is correct The correct one is $\Leftrightarrow y(a-b)(a+b-2c)[s_a^{k-1}+s_a^{k-2}s_b+...+s_b^{k-1}]\ge \\x(b-c)(b+c-2a)[s_b^{k-1}+s_b^{k-2}s_c+...+s_c^{k-1}]$ $\Leftrightarrow (x+2y)[s_a^{k-1}+s_a^{k-2}s_b+...+s_b^{k-1}]\ge \\ -(2x+y)[s_b^{k-1}+s_b^{k-2}s_c+...+s_c^{k-1}]$ which is trivial EDIT in fact $ys_a^k+xs_c^k\ge (x+y)s_b^k$ is correct since $s_a, \ s_c>s_b$

MathUniverse wrote: This inequality can be proved using induction and global derivative in few lines, but that makes problem not suitable for olympiad. I'd like to know what the official solution is. Proof: We will prove inequality for all $a,b,c\ge 0$ and $k \in \mathbb{N}$ $\sum (a^2+2bc)^k \leq (a^2+b^2+c^2)^k+2(ab+ac+bc)^k \quad (1)$. If $k=1$, we have equality. Now suppose it holds for some $k-1$. Now, we define $F(a,b,c)=RHS-LHS$ in $(1)$. If we denote global deribvative with $[F]$, then: $[F]=2k(a+b+c)\left[(a^2+b^2+c^2)^{k-1}+2(ab+ac+bc)^{k-1}-\sum (a^2+2bc)^{k-1} \right]$ Because of assumption, we have $[F]\ge 0$. Now, using global derivative theorem, it is enough to prove inequality when $abc=0$. WLOG we can suppose $c=0$, so it remains to prove: $a^{2k}+b^{2k}+(2ab)^{k}\le (a^2+b^2)^k+2(ab)^k$ $\Leftrightarrow (2^k-2)(ab)^k \le \sum_{i=1}^{k-1} \dbinom{k}{i}a^{2i}b^{2(k-i)}$ What follows from Am-Gm. Equality holds when $a=b=c$ and when $a=b, \; c=0$ with all permutations. $\blacksquare$ My idea is the same as yours,when I glanced at this problem.But actully we have a elementary version of the proof,that is ,Denoting \[f\left( {a,b,c} \right) = {\left( {{a^2} + {b^2} + {c^2}} \right)^n} + 2{\left( {ab + bc + ca} \right)^n} - \sum {{{\left( {{a^2} + bc} \right)}^n}} ,\]and for $t \ge - \min \left\{ {a,b,c} \right\}$,show $g'\left( t \right) = {f_t}^\prime \left( {a + t,b + t,c + t} \right) \ge 0$ with indution hypothesis.

MathUniverse wrote: We will prove inequality for all $a,b,c\ge 0$ and $k \in \mathbb{N}$ $\sum (a^2+2bc)^k \leq (a^2+b^2+c^2)^k+2(ab+ac+bc)^k$ For $k=0$, the inequality becomes $$(a^2+b^2+c^2)(ab+bc+ca)^2\ge (a^2+2bc)(b^2+2ca)(c^2+2ab)$$which follows immediately from the identity $$(a^2+b^2+c^2)(ab+bc+ca)^2=(a^2+2bc)(b^2+2ca)(c^2+2ab)+(a-b)^2(b-c)^2(c-a)^2$$Is the following inequality true: $$\sqrt{a^2+b^2+c^2}+2\sqrt{ab+bc+ca}\ge \sqrt{a^2+2bc}+\sqrt{b^2+2ca}+\sqrt{c^2+2ab}$$

See here: https://artofproblemsolving.com/community/c6h345854p1850961

Also see here: https://artofproblemsolving.com/community/c6h549887 Take $(a, b, c)\mapsto (a^2+b^2+c^2, ab+bc+ca, ab+bc+ca)$, etc.

Let $s_n = RHS$, $t_n = LHS$. Note that the inequality is an equality for $n = 0,1,2$. Moreover, $$s_n = \sigma_1s_{n-1} - \sigma_2s_{n-2} + \sigma_3s_{n-3}$$$$t_n = \tau_1s_{n-1} - \tau_2s_{n-2} + \tau_3s_{n-3}$$where the $\sigma_i$ and $\tau_i$ are symmetric sums in $(a^2+b^2+c^2, ab+bc+ca, ab+bc+ca)$ and $(a^2 + bc)_{cyc}$ respectively; $\sigma_1 = \tau_1, \sigma_2 = \tau_2$. Let $d_n = s_n - t_n$, then $d_0 = d_1 = d_2 = 0$, $$d_n = \sigma_1d_{n-1} - \sigma_2d_{n-2} + \sigma_3s_{n-3} - \tau_3t_{n-3}.$$Note that $d_3 \ge 0 \iff \sigma_3 \ge \tau_3$, and we can show $d_3 \ge 0$ after expanding both sides. Having shown that $d_3 \ge 0$ and $\sigma_3 \ge \tau_3$, we have $$d_n \ge \sigma_1d_{n-1} - \sigma_2d_{n-2} + \sigma_3d_{n-3}.$$ Consider the following process: whenever we have an inequality of the form $d_n \ge \sigma^{(i)}_1 d_{n-1-i} - \sigma^{(i)}_2d_{n-2-i} + \sigma^{(i)}_3d_{n-3-i}$ and $\sigma_1^{(i)} \ge 0$, we will use the estimate $d_{n-1-i} \ge \sigma_1d_{n-2-i} - \sigma_2d_{n-3-i} + \sigma_3d_{n-4-i}$ to obtain a new inequality $$d_n \ge \sigma_1^{(i)}(\sigma_1d_{n-2-i} - \sigma_2d_{n-3-i} + \sigma_3d_{n-4-i}) - \sigma_2^{(i)}d_{n-2-i} + \sigma_3^{(i)}d_{n-3-i}\\ =: \sigma_1^{(i+1)}d_{n-1-(i+1)} - \sigma_2^{(i+1)}d_{n-2-(i+1)} + \sigma_3^{(i+1)}d_{n-3-(i+1)}. $$If we are able to repeat this until $i = n-2$ then we are done, so it suffices to show that $\sigma_1^{(i)}$ remains nonnegative throughout the process. Remark that $$\sigma_1^{(i+1)} = \sigma_1^{(i)}\sigma_1 - \sigma_2^{(i)}\\ \sigma_2^{(i+1)} = \sigma_1^{(i)}\sigma_2 - \sigma_3^{(i)}\\ \sigma_3^{(i+1)} = \sigma_3\sigma_1^{(i)} $$which we can use to solve for the sequence $(\sigma_1^{(m)})_{m\ge 0}$ and obtain $$\sigma_1^{(i+1)} = \sigma_1^{(i)}\sigma_1 - \sigma_1^{(i-1)}\sigma_2 + \sigma_3\sigma_1^{(i-2)}\quad \forall \ i \ge 2.$$This is precisely the recursion for $d_n$ with inequality replaced by equality; the initial terms of the sequence are $(\sigma_1, \sigma_1^2 - \sigma_2, \sigma_1^3 - 2\sigma_1\sigma_2 + \sigma_3)$, and we can back-track to get $\sigma_1^{(-1)} = 1$. Now if we let $(x,y,z) = (a^2+b^2+c^2, ab+bc+ca, ab+bc+ca)$ then we claim that $$\sigma^{(i-1)}_1 = \sum_{j+k+l = i} x^jy^kz^l$$which is nonnegative because $x,y,z \ge 0$. The proof is by induction on $i$, counting the number of times $x^jy^kz^l$ appear on both sides of the equation. Therefore $\sigma_1^{(i)}$ remains nonnegative in the process above, so we are done.

Let $$S_k=(a^2+2bc)^k + (b^2+2ac)^k + (c^2+2ab)^k$$ $$T_k= (a^2+b^2+c^2)^k + 2(ab+bc+ca)^k$$ Be Newton Sums. We can check that $$S_0=T_0=3$$ $$S_1=T_1=(a+b+c)^2$$ $$S_2=T_2=\sum_{cyc} (a^4+4a^2b^2 + 4a^2bc)$$ Let $$P(x)=(x-(a^2+2bc))(x-(b^2+2ac))(x-(c^2+2ab))$$ $$Q(x)=(x-(a^2+b^2+c^2))(x-(ab+bc+ca))^2$$ We have $[x]P(x)= \frac{S_1^2 - S_2}{2} = \frac{T_1^2-T_2}{2} = [x]Q(x)$ Therefore, $P(x)-Q(x) = (a^2+b^2+c^2)(ab+bc+ca)^2 - \prod\limits_{cyc} (a^2+2bc)$ Claim: $(a^2+b^2+c^2)(ab+bc+ca)^2 \le \prod\limits_{cyc} (a^2+2bc)$ Proof: Note $(a^2+b^2+c^2, ab+bc+ca, ab+bc+ca) \succ (a^2+2bc, b^2+2ac, c^2+2ab)$ for any ordering of $(a^2+2bc, b^2+2ac, c^2+2ab)$. By Karamata's inequality that if $f(x)$ is concave and we have $x_1+x_2+x_3=y_1+y_2+y_3, x_1\ge x_2\ge x_3, y_1\ge y_2\ge y_3$, then $f(x_1)+f(x_2)+f(x_3) \le f(y_1)+f(y_2)+f(y_3)$. The claim follows by setting $f(x)=\log x, x_1=a^2+b^2+c^2, x_2=x_3=ab+bc+ca, $ and $y_1,y_2,y_3$ is $a^2+2bc, b^2+2ac, c^2+2ab$ in non-increasing order. Thus, let $P(x)=x^3-Sx^2+Tx-p, Q(x)=x^3-Sx^2+Tx-q$ then $p\ge q$. We induct on $k$ to show that $S_k\le T_k$. The base cases, $k\in \{0,1,2\}$ has been dealt with earlier. We have $S_k=SS_{k-1} - TS_{k-2} + pS_{k-3} \le ST_{k-1}-TT_{k-2}+qT_{k-3}=T_k$, completing the induction.

bobthesmartypants wrote: Also see here: https://artofproblemsolving.com/community/c6h549887 Take $(a, b, c)\mapsto (a^2+b^2+c^2, ab+bc+ca, ab+bc+ca)$, etc. indeed