The inequality is homogenous, so WLOG $abc=1$; then $a+b+c=4.$ WLOG $a$ is the smallest of the three variables. Then we want to show that $$2ab+2bc+2ac\ge b^2+c^2-3a^2$$$$(a+b+c)^2\ge 2b^2+2c^2-2a^2$$$$b^2+c^2-a^2\le 8$$$$b^2+c^2-(4-b-c)^2\le 8$$$$8b+8c-2bc\le 24$$$$4(4-a)-\frac{1}{a}\le 12$$$$4a-4+\frac{1}{a}\ge 0$$$$(2a-1)^2\ge 0$$which is true.

$\frac{4a\sqrt[3]{abc}+bc}{2}\geq \sqrt{4(abc)^{\frac{4}{3}}}$ by AM-GM, and substituting $4\sqrt[3]{abc}=a+b+c$ gives $4a(a+b+c)+4bc\geq (a+b+c)^2$ so $2(ab+bc+ca)+4a^2\geq a^2+b^2+c^2$ as desired.

Without loss of generality, let $abc=1$ and let $\text{min}(a^2,b^2,c^2)=a^2$. This implies $a+b+c=4$. $2(ab+ac+bc) + 4\text{min}(a^2,b^2,c^2) \ge a^2+b^2+c^2$ $\iff 2(ab+ac)+ 2bc + 4a^2 \ge a^2+b^2+c^2$ $\iff 2a(4-a)+3a^2 \ge (b-c)^2$. Note that $(b-c)^2=(b+c)^2-4bc=(4-a)^2-\frac{4}{a}$. Then we need to prove $2a(4-a)+3a^2 \ge (4-a)^2-\frac{4}{a} \iff 4a^2-4a+1 \ge 0$, which is true. Equality holds for $a:b:c=1:\frac{7-\sqrt{17}}{2}: \frac{7+\sqrt{17}}{2}$ and its permutations.

WLOG let $c = \min(a,b,c) = 1$ by scaling. The given inequality becomes equivalent to \[ 4ab + 2a + 2b + 3 \ge (a+b)^2 \qquad \forall a+b = 4(ab)^{1/3}-1. \]Now, let $t = (ab)^{1/3}$ and eliminate $a+b$ using the condition, to get \[ 4t^3 + 2(4t-1) + 3 \ge (4t-1)^2 \iff 0 \le 4t^3 - 16t^2 + 16t = 4t(t-2)^2 \]which solves the problem. Equality occurs only if $t=2$, meaning $ab = 8$ and $a+b=7$, which gives \[ \{a,b\} = \left\{ \frac{7 \pm \sqrt{17}}{2} \right\} \]with the assumption $c = 1$. Scaling gives the curve of equality cases.

Mt. Inequality has exploded a second time

v_Enhance wrote: WLOG let $c = \min(a,b,c) = 1$ by scaling. The given inequality becomes equivalent to \[ 4ab + 2a + 2b + 3 \ge (a+b)^2 \qquad \forall a+b = 4(ab)^{1/3}-1. \]Now, let $t = (ab)^{1/3}$ and eliminate $a+b$ using the condition, to get \[ 4t^3 + 2(4t-1) + 3 \ge (4t-1)^2 \iff 0 \le 4t^3 - 16t^2 + 16t = 4t(t-2)^2 \]which solves the problem. Equality occurs only if $t=2$, meaning $ab = 8$ and $a+b=7$, which gives \[ \{a,b\} = \left\{ \frac{7 \pm \sqrt{17}}{2} \right\} \]with the assumption $c = 1$. Scaling gives the curve of equality cases. wait... plz tell me we didn't have to give equality cases

Do you lose points if you don't state the equality case?

Mountain Inequality just got an aftershock! well, just normalize the inequality as $abc = 1$ then we have $a+b+c = 4$ note that the inequality we want is $4\sum_{cyc} ab + 4min(a^2,b^2,c^2) \geq (a+b+c)^2 = 16$ or, $ab + bc + ca + min(a^2,b^2,c^2) \geq 4$ , let $c = min(a,b,c)$ Then, $c^2 = min(a^2,b^2,c^2)$ since $ab = \frac{1}{c}, bc+ca = (c)(a+b) = 4c-c^2$ all we left to prove is $\frac{1}{c} + 4c-c^2 +c^2 \geq 4$ or $\frac{1}{c} + 4c \geq 4$ which is true from AM-GM

The_Turtle wrote: Do you lose points if you don't state the equality case? they never asked for it right, so i think it should be fine

No, it does not ask for the equality case but it does break ties

Let $a+b+c = 4k, abc = k^3$ and put everything in terms of $k,a$. You get $\frac{k}{a} (2a - k)^2 \ge a^2 - min(a^2, b^2, c^2)$ and you are done.

Does this work? Please tell me if it doesn't. Since the inequality as well as the given condition is homogenous, let $a+b+c=4\sqrt[3]{abc}=1.$ Then $a+b=1-c,ab=\frac{1}{64c}.$ Further stipulate $a\geq b\geq c$ to break the symmetry. Now the inequality is equivalent to $$2ab+2bc+2ca+4c^2\geq a^2+b^2+c^2$$$$\iff 4(c+a)(c+b)\geq (a+b+c)^2$$$$\iff c(a+b+c)+ab\geq\frac{1}{4}$$$$\iff c+\frac{1}{64c}\geq \frac{1}{4}$$$$\iff c^2-\frac{c}{4}+\frac{1}{64}\geq 0$$$$\iff (c-1/8)^2\geq 0$$

pandadude wrote: No, it does not ask for the equality case but it does break ties this should be false... if they don't ask for something, then it's not necessary... memorable tiebreakers have been listing subsets on 2016 U1, and apparently grade on 2016 USAJMO MOP qualification.

Oops does this work.

@above that's along the lines of what i did -- i think it's fine.

Solution attached

RedFireTruck wrote: WLOG $a\le b\le c$. We can add $2(ab+bc+ca)$ to both sides to get $4(a^2+ab+bc+ca)\ge (a+b+c)^2=16a^\frac23b^\frac23c^\frac23$. This simplifies further into $a(a+b+c)+bc=4a^\frac43b^\frac13c^\frac13+bc\ge 4a^\frac23b^\frac23c^\frac23$. We can directly apply AM-GM to see this is true. ok i kinda forgot that i did this problem so i like resolved it but i think my new solution is worse oops since the condition and inequality are both homogenous, let $a+b+c=4$ and $abc=1$ then, we wanna prove $4(ab+bc+ca)+4\min(a^2, b^2, c^2)\ge 16$ since like $ab+bc+ca+a^2=(a+b)(a+c)=(4-c)(4-b)$, we wanna prove $(4-a)(4-b)(4-c)\ge 4(4-\min(a,b,c))$ expanding gives $(4-a)(4-b)(4-c)=64-16(a+b+c)+4(ab+bc+ca)-abc=4(ab+bc+ca)-1$ so we wanna prove $ab+bc+ca+\min(a,b,c)\ge\frac{17}{4}$ wlog let $a\le b\le c$ so then we wanna prove $a(1+b+c)+bc=a(5-a)+\frac1a\ge\frac{17}{4}$ cuz $a+b+c=4$ and $abc=1$ then we just gotta show that $-4a^3+20a^2-17a+4\ge 0$ for all $0<a\le \frac43$ when $a=0$ we get $4\ge0$ and when $a=2$ we get $-32+80-34+4\ge0$ which are both true taking the derivative we get $-12a^2+40a-17=0$ or $(6a-10)^2=49$ so the derivative is $0$ at $a=\frac12$ and $a=\frac{17}{6}$ at $a=\frac12$, we get a local minimum at $-\frac12+5-\frac{17}{2}+4=0$ and $\frac{17}{6}>2$, so $-4a^3+20a^2-17a+4\ge 0$ for all $0<a\le \frac43$, as desired.

Without loss of generality $abc = 1$ so $a + b + c = 4$. Then, \begin{align*} 2(ab + bc + ca) + 4 \min(a^2, b^2, c^2) &\geq a^2 + b^2 + c^2\\ \iff 4(ab + bc + ca) + 4 \min(a^2, b^2, c^2) &\geq 16\\ \iff ab + bc + ca + \min(a^2, b^2, c^2) &\geq 4\\ \end{align*}Now without loss of generality $\min(a^2, b^2, c^2) = a^2$ so that, \begin{align*} ab + bc + ca + \min(a^2, b^2, c^2) &\geq 4\\ \iff a(b + c) + bc + a^2 &\geq 4\\ \iff a(4 - a) + \frac{1}{a} + a^2 &\geq 4\\ \iff 4a + \frac{1}{a} &\geq 4\\ \end{align*}which follows by AM-GM. Oops, looks like I did this before.

(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca=16(abc)^2/3 16(abc)^2/3-(a^2+b^2+c^2)+4min(a^2,b^2,c^2)>=a^2+b^2+c^2 8(abc)^2/3-(a^2+b^2+c^2)+2min(a^2,b^2,c^2)>=0 8(abc)^2/3-(b^2+c^2)+a^2>=0 2(a^2+b^2+c^2)-(b^2+c^2)+a^2>=0 3a^2+b^2+c^2>=0 Hence proved

Can someone explain the title bruh

WLOG $abc=1, a+b+c=4$ because everything is homogeneous. Also WLOG $a \geq b \geq c$, so we have $$2(ab+bc+ac) + 4c^2 \geq a^2+b^2+c^2$$$$2(ab+bc+ac)+4c^2 \geq 16-2(ab+bc+ac)$$$$ab+bc+ac+c^2 \geq 4$$$$\frac{1}{c} + c(4-c) + c^2 \geq 4$$$$\frac{(c - \frac12)^2}{c} \geq 0$$, trivial.

By WLOG and scaling, let $1=a\le b\le c$. We know that $1+b+c=4\sqrt[3]{bc}$ and want to prove that $$2(b+c+bc)+4\ge 1+b^2+c^2.$$If we let $x=\sqrt[3]{bc}$, we know that $b+c=4x-1$ and $bc=x^3$. We want to prove that $$2(4x-1+bc)+3\ge (b+c)^2-2bc$$$$=(4x-1)^2-2bc$$$$=16x^2-8x+1-2bc$$$$=16x^2-8x+1-2x^3.$$Simplifying, we want to show that $$2(4x-1+x^3)+3\ge 16x^2-8x+1-2x^3,$$or $$4x^3-16x^2+16x\ge 0.$$This factors as $$4x(x-2)^2\ge 0,$$which is true.

Ooooh, cute problem. I really liked this even though it's an inequality. Since the inequality and condition, are both symmetric in $a,b$ and $c$. Without loss of generality assume $a\ge b \ge c$. Note that it suffices to show, \begin{align*} 2(ab+bc+ca)+4c^2 &\ge a^2+b^2 + c^2 \\ 2(ab+bc+ca) + 4c^2&\ge (a+b+c)^2 - 2(ab+bc+ca)\\ 4(ab+bc+ca)+4c^2 & \ge 16(abc)^{\frac{2}{3}}\\ ab+bc+ca+c^2 & \ge 4(abc)^{\frac{2}{3}} \end{align*}But, this is actually quite clear since by the AM-GM Inequality, \begin{align*} ab+bc+ca+c^2 &= ab + c(a+b+c)\\ & = ab + 4c(abc)^\frac{1}{3}\\ & \ge 2\sqrt{4(abc)^\frac{4}{3}}\\ &= 4(abc)^{\frac{2}{3}} \end{align*}as desired.

Because the inequality is homogeneous, assume that WLOG $abc = 1 \implies a + b + c = 4$. Also, assume that WLOG, $a \leq b \leq c$. We then want to show that \[2(ab + bc + ca) + 4a^2 \geq a^2 + b^2 + c^2\] We now add $2(ab + bc + ca)$ to both sides of the inequality; we then obtain \[4(ab + bc + ca) + 4a^2 \geq a^2 + b^2 + c^2 + 2(ab + bc +ca)\] Recall that $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc +ca)$. Thus, our inequality simplifies to \[4(ab + bc + ca) + 4a^2 \geq (a + b + c)^2\]\[\implies 4(ab + bc + ca) + 4a^2 \geq 16\]\[\implies ab + bc + ca + a^2 \geq 4\] We now turn the LHS into an expression that consists of only $a$ and constants. Note that $b + c = 4 - a$ and $bc = \frac{1}{a}$; thus, our inequality is \[a(b + c) + bc + a^2 \geq 4\]\[a(4 - a) + \frac{1}{a} + a^2 \geq 4\]\[4a + \frac{1}{a} \geq 4\]\[4a^2 - 4a + 1 \geq 0\]\[(2a - 1)^2 \geq 0\] which is true by the trivial inequality, as desired.

secret technique [s c h i z o p h r e n i a | b l a s t] WLOG $a \le b \le c$. Then, the givens rewrite to \[ 4ab + 4bc + 4ca + 4a^2 \ge (a + b + c)^2. \]We impose the condition $a + b + c = 4$, whence $abc = 1$ since the condition and inequality are homogenous. Therefore it is sufficient to prove \[ (a + b)(a + c) \ge 4,\]or \[ (4 - c)(4 - b) \ge 4.\]Make the substitution $p = 4-a$ and cyclic. We are given the conditions \[p + q + r = 8 \]and \[ (4-p)(4-q)(4-r) = 1\]and we wish to show that \[ qr \ge 4. \]Let $qr = s$. Then, we have \[ 64 - 16p - 16q - 16r + 4pq + 4qr + 4rp - pqr = 1 \]so \[ 4p(q + r) + qr(4 - p) = 65. \]Working more with this equation, \begin{align*} 4p(8 - p) + s(4 - p) &= 65 \\ 4p^2 - 32p - 4s + sp + 65 &= 0 \\ 4p^2 + (s-32)p + 65 - 4s &= 0. \end{align*}Now the discriminant says that \[ (s-32)^2 - 16(65 - 4s) \ge 0, \]so \[ s^2 - 64s + 32^2 + 64s - 16 \cdot 65 \ge 0\]and thus \[ s^2 \ge 16.\]It follows then that since $s > 0$ we have $s \ge 4$. This is what we want.

Please contact westskigamer@gmail.com if there is an error with my solution.

Here is my approach. Let $a$ be their minimum. Then $b=ma$ and $ c=na $ for some $m \geq 1$ , and $n \geq1 $. Then we have $1+m+n=4\sqrt[3]{mn}$ and since our statement is equivalent to $4(ab+ac+bc)+4a^2 \geq (a+b+c)^2$ which is $4(1+m)(1+n) \geq (1+m+n)^2$. If we expand and clear the terms we will have $3+2(m+n)+4mn \geq (m+n)^2$ and combining it with $1+m+n=4\sqrt[3]{mn}$ and let $ x=m+n$ and $y=mn$, then we have $f(x)=x^3-13x^2+35x+49 \geq 0$ for $x>2$. And done since it's true for sure.

We will first prove unconditionally that \[(a+c)(b+c)\ge 2\sqrt{abc(a+b+c)}.\]Indeed, if we square both sides, the inequality is actually equivalent to $(ab-ac-bc-c^2)^2\ge 0$. It is not hard to see that if $c=\min(a,b,c)$, then the thing we proved is equivalent to the desired result.