This was USAJMO #2, not 3.

dang. that is pretty tricky numbering. anyways

I found $(a+b)^2 + (b+c)^2 + (c+a)^2 \le 4$, so I said that $(a+b)^2 \le 4$ and the others too, $\dfrac {1}{(a+b)^2} \ge \dfrac {1}{4}$ $\dfrac {ab+1}{(a+b)^2} \ge \dfrac {ab+1}{4}$ So it becomes $\dfrac {ab+bc+ac+ 3}{4} \ge 3$ So I was trying to prove that $ab + bc + ac \ge 9$, but it wasn't working (because it isn't true). So yeah, I guess I failed. The inequality was too weak?

Math Geek wrote: Let $x=a+b$, $y=b+c$, $z=a+c$. I was so ridiculous. I contemplated doing that, and decided not to ONLY because I thought the expressions for $ab,bc,ca$ would be ugly!

Yongyi781 wrote: Math Geek wrote: Let $x=a+b$, $y=b+c$, $z=a+c$. I was so ridiculous. I contemplated doing that, and decided not to ONLY because I thought the expressions for $ab,bc,ca$ would be ugly! Same! I ended up removing the 1 from the numerator and using AM-GM; but I still didn't prove it completely.

I am probably the only one silly enough to know this majorization lemma thingy but not how to factor. EDIT: which apparently doesn't work anyway .

I never got to the end of this one, but I somehow used AM-GM twice and ended up trying to prove this: When $(a+b)^2(b+c)^2(a+c)^2 < \left(\frac43\right)^3$, it will never exceed $(ab+1)(bc+1)(ac+1)$.

On a side note, what I tried to do was this: \begin{align*}4\sum\frac{ab+1}{(a+b)^2}&\geq\sum\frac{ab+1}{(a+b)^2}\sum(ab+1)\\ &\geq\left(\sum\frac{ab+1}{a+b}\right)^2\\ &\geq 12\\ \iff\sum\frac{ab+1}{a+b}&\geq 2\sqrt3.\end{align*}As far as I know, that last inequality is also true, but it seems as if it's actually much harder to prove than the original inequality. So much for 4 hours...

you use the substitution x=a+b y=b+c z=c+a and you can reduce the inequality to x^3y^3+x^3z^3+y^3z^3<=3x^2y^2z^2 from this point you can either use AM-GM or muirhead, guess which one I used on the test?

So I did a different substitution than most!

Yongyi781 wrote: On a side note, what I tried to do was this: ** what 1=2 did** As far as I know, that last inequality is also true, but it seems as if it's actually much harder to prove than the original inequality. So much for 4 hours... high five =P

I used triangle notation, homogenized, multiplied everything out, then used two applications of Muirhead.

I took the original given inequality, expanded, simplified to get : $a^2+b^2+c^2+ab+bc+ac<=2$. Then I took three cases, added ab, bc, and ac to each. Then I added all the equations, used AM-GM,...and solved it. Quick Question: For example, when i added ab, bc, and ac, do I have to show work for all three? Or can I just show one, and say similarly ...

Would we be docked if we didn't explicitly state when equality occurs?

giratina150 wrote: Quick Question: For example, when i added ab, bc, and ac, do I have to show work for all three? Or can I just show one, and say similarly ... Saying "similarly" should be fine, because everything's symmetric. cwein3 wrote: Would we be docked if we didn't explicitly state when equality occurs? I'm pretty sure not. The problem itself didn't ask to state when equality occurs.

Does anyone know how to solve my "harder" version of the problem, by the way?

I don't know if it is true. Plug in $a=b=c=\frac{1}{\sqrt{3}}$.

my otis app (showed ortogonol a while ago) storage

Note that $1\geq \frac{a^2+b^2+c^2+(a+b+c)^2}{4}$. Let us now homogenize this inequality, meaning that the original inequality is greater than or equal to: \begin{align*} \sum_\text{cyc}{\frac{ab}{(a+b)^2}}+\sum_\text{cyc}{\frac{a^2+b^2+c^2+(a+b+c)^2}{4(a+b)^2}}&=\sum_\text{cyc}{\frac{ab}{(a+b)^2}}+\sum_\text{cyc}{\frac{(a+b)^2+(b+c)^2+(a+c)^2}{4(a+b)^2}}\\ &=\sum_\text{cyc}{\frac{(b+c)^2+(c+a)^2+4ab}{4(a+b)^2}}+\frac{3}{4} \end{align*} Without loss of generality, assume that $a\geq b\geq c$. Therefore, by AM-GM: \begin{align*} \sum_\text{cyc}{\frac{(b+c)^2+(c+a)^2+4ab}{4(a+b)^2}}+\frac{3}{4}&\geq \sum_\text{cyc}{\frac{(2c)^2+(2c)^2+4c^2}{4(2a)^2}}+\frac{3}{4}\\ &=\sum_\text{cyc} \left(\frac{3}{4}\cdot \frac{c^2}{a^2}\right ) +\frac{3}{4}\\ &\geq \frac{3}{4}\cdot 3+\frac{3}{4}\\ &=3,\\ \end{align*}as desired.

Note that all $\sum,\prod$ notations used here are cyclic and not symmetric. Clearly, we have, $1 \geq \frac{\left(\sum{a^2}+\sum{ab}\right)}{2}$ Now $\sum{\frac{ab+1}{\left(a+b\right)^2} \geq \sum{\frac{ab+ \frac{\left(\sum{a^2}+\sum{ab}\right)}{2}}{\left(a+b\right)^2}}} \geq \sum{\left(\frac{1}{2}+\frac{c^2+ac+bc+ab}{\left(a+b\right)^2}\right)}={\frac{3}{2}+\sum{\frac{(c+a)(c+b)}{2\left(a+b\right)^2}}}$ Notice that, by AM-GM,$\sum{\frac{(c+a)(c+b)}{2\left(a+b\right)^2}} \geq \frac{3}{2}$ This gives us the conclusion that, $\frac{3}{2}+\sum{\frac{(c+a)(c+b)}{2\left(a+b\right)^2}} \geq \frac{3}{2}+\frac{3}{2} = 3$. Our proof is thus complete. Case of equality: Because we used AM-GM, we notice the case of equality holds $\iff \frac{(c+a)(c+b)}{2(a+b)^2}=\frac{(a+b)(a+c)}{2(b+c)^2}$ $\iff (b+c)^3=(a+b)^3$ $\iff b=c$, Now we can cyclically obtain other expressions to have $a=b=c$. In the other given inequality, we can plug in $a=b=c=k$ to have $3k^2+9k^2 \leq 4$. As $k \in \mathbb{R}^{+}$, equality here holds $\iff k=\frac{1}{\sqrt3}$. Thus for equality, we should have $a=b=c=\frac{1}{\sqrt3}$

Let $x=a+b,$ $y=b+c,$ and $z=a+c.$ Note that $a=\dfrac{x+z-y}{2},$ $b=\dfrac{x+y-z}{2},$ $c=\dfrac{y+z-x}{2},$ and $x^2+y^2+z^2=(a+b)^2+(b+c)^2+(a+c)^2=2a^2+2b^2+2c^2+2ab+2bc+2ac=a^2+b^2+c^2+(a+b+c)^2\leq 4.$ Since $x^2+y^2+z^2>0,$ we have $\dfrac{1}{4}\leq \dfrac{1}{x^2+y^2+z^2},$ so $\dfrac{ab+1}{(a+b)^2}+\dfrac{bc+1}{(b+c)^2} +\dfrac{ac+1}{(c+a)^2}=\dfrac{\left(\dfrac{x+z-y}{2}\right)\left(\dfrac{x+y-z}{2}\right)+1}{x^2}+\dfrac{\left(\dfrac{x+y-z}{2}\right)\left(\dfrac{y+z-x}{2}\right)+1}{y^2}+\dfrac{\left(\dfrac{y+z-x}{2}\right)\left(\dfrac{x+z-y}{2}\right)+1}{z^2}=\dfrac{\left(\dfrac{x^2}{4}-\dfrac{y^2}{4}-\dfrac{z^2}{4} +\dfrac{yz}{2}+1\right)}{x^2}+\dfrac{\left(\dfrac{y^2}{4}-\dfrac{x^2}{4}-\dfrac{z^2}{4}+\dfrac{xz}{2}+1\right)}{y^2}+\dfrac{\left(\dfrac{z^2}{4}-\dfrac{x^2}{4}-\dfrac{y^2}{4}+\dfrac{xy}{2}+1\right)}{z^2}=\dfrac{3}{4}-\dfrac{y^2}{4x^2}-\dfrac{x^2}{4y^2}-\dfrac{y^2}{4z^2}-\dfrac{z^2}{4y^2}-\dfrac{x^2}{4z^2}-\dfrac{z^2}{4x^2}+\dfrac{yz}{2x^2}+\dfrac{xz}{2y^2}+\dfrac{xy}{2z^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}.$ We now see the desired result follows if the following inequality is true: $\dfrac{yz}{2x^2}+\dfrac{xz}{2y^2}+\dfrac{xy}{2z^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\geq \dfrac{9}{4}+\dfrac{y^2}{4x^2}+\dfrac{x^2}{4y^2}+\dfrac{y^2}{4z^2}+\dfrac{z^2}{4y^2}+\dfrac{x^2}{4z^2}+\dfrac{z^2}{4x^2}.$ Now, note that $\dfrac{yz}{2x^2}+\dfrac{xz}{2y^2}+\dfrac{xy}{2z^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=\dfrac{yz}{2x^2}+\dfrac{xz}{2y^2}+\dfrac{xy}{2z^2}+\dfrac{\left(x^2+y^2+z^2\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)}{x^2+y^2+z^2}\geq \dfrac{yz}{2x^2}+\dfrac{xz}{2y^2}+\dfrac{xy}{2z^2}+\dfrac{\left(x^2+y^2+z^2\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)}{4}=\dfrac{yz}{2x^2}+\dfrac{xz}{2y^2}+\dfrac{xy}{2z^2}+\dfrac{\left(3+\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+\dfrac{z^2}{y^2}+\dfrac{y^2}{z^2}+\dfrac{x^2}{z^2}+\dfrac{z^2}{x^2}\right)}{4}=\dfrac{yz}{2x^2}+\dfrac{xz}{2y^2}+\dfrac{xy}{2z^2}+\dfrac{3}{4}+\dfrac{x^2}{4y^2}+\dfrac{y^2}{4x^2}+\dfrac{z^2}{4y^2}+\dfrac{y^2}{4z^2}+\dfrac{x^2}{4z^2}+\dfrac{z^2}{4x^2}.$ By AM-GM, $\dfrac{\left(\dfrac{xy}{z^2}+\dfrac{xz}{y^2}+\dfrac{yz}{x^2}\right)}{3}\geq \sqrt[3]{\dfrac{xy}{z^2}\cdot\dfrac{xz}{y^2}\cdot\dfrac{yz}{x^2}}=1,$ so $\dfrac{xy}{2z^2}+\dfrac{yz}{2x^2}+\dfrac{xz}{2y^2}\geq \dfrac{3}{2}.$ As such, $\dfrac{yz}{2x^2}+\dfrac{xz}{2y^2}+\dfrac{xy}{2z^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\geq \dfrac{yz}{2x^2}+\dfrac{xz}{2y^2}+\dfrac{xy}{2z^2}+\dfrac{3}{4}+\dfrac{x^2}{4y^2}+\dfrac{y^2}{4x^2}+\dfrac{x^2}{4z^2}+\dfrac{z^2}{4x^2}+\dfrac{y^2}{4z^2}+\dfrac{z^2}{4y^2}\geq \dfrac{9}{4}+\dfrac{x^2}{4y^2}+\dfrac{y^2}{4x^2}+\dfrac{x^2}{4z^2}+\dfrac{z^2}{4x^2}+\dfrac{y^2}{4z^2}+\dfrac{z^2}{4y^2},$ and the desired result follows. $\Box$

Let $a+b = 2x, b + c =2 y, a + c = 2z$ then we have $x^2 + y^2 + z^2 \le 1$ Plugging these in the desired inequality gives $$\sum\frac{x^2 - (y - z)^2 + 1}{4x^2} \ge 3 \iff [\sum\frac{-(y - z)^2 + 1}{x^2} \ge 9 $$Multiplying by $x^2y^2z^2$ gives $$\sum -y^4z^2 - y^2z^4 + 2y^3z^3 + y^2z^2 \ge \sum -y^4z^2 - y^2z^4 + 2y^3z^3 + y^2z^2(x^2 + y^2 + z^2)$$$$= \sum x^2y^2z^2 + 2y^3z^3 \ge 9x^2y^2z^2$$$$\iff \sum 2y^3z^3 \ge 6x^2y^2z^2$$which is true by Muirhead's Inequality.

it is easy to prove $a^2+b^2+c^2+(a+b+c)^2=2a^2+2b^2+2c^2+2ab+2bc+2ca=(a+b)^2+(b+c)^2+(c+a)^2 \leq 4$ also, by AM-GM, $2 \geq a^2+b^2+c^2+ab+bc+ca \geq 2ab+2bc+2ca$, and $1 \geq ab+bc+ca$, $4 \geq (ab+1)+(bc+1)+(ca+1)$ it is obvious that if $x_1+y_1+z_1 \leq n$ and $x_2+y_2+z_2 \leq n$, $\frac{x_1}{x_2}+\frac{y_1}{y_2}+\frac{z_1}{z_2} \geq 3$ thus, $\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ca+1}{(c+a)^2}\geq 3$

The condition $a^2+b^2+c^2+(a+b+c)^2\leq4$ rewrites as $(a+b)^2+(b+c)^2+(c+a)^2\le 4$. By AM-HM, $$\frac{3}{\frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}}\le \frac{(a+b)^2+(b+c)^2+(c+a)^2}{3}\le \frac{4}{3}$$so $$\frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}\ge \frac94.$$ Now it suffices to prove that $\frac{ab}{(a+b)^2}+\frac{bc}{(b+c)^2}+\frac{ca}{(c+a)^2}\ge \frac34$. By Cauchy-Schwarz and AM-GM, $$3(\frac{ab}{(a+b)^2}+\frac{bc}{(b+c)^2}+\frac{ca}{(c+a)^2})\ge (\frac{\sqrt{ab}}{a+b}+\frac{\sqrt{bc}}{b+c}+\frac{\sqrt{ca}}{c+a})^2\ge (\frac32)^2=\frac94$$so $$\frac{ab}{(a+b)^2}+\frac{bc}{(b+c)^2}+\frac{ca}{(c+a)^2}\ge \frac34$$as desired.

$$a^2 + b^2 + c^2 + (a + b + c)^2 \le 4 \Longleftrightarrow a^2 + b^2 + c^2 + ab + bc + ca \le 2 \Longleftrightarrow (a + b)^2 + (b + c)^2 + (c + a)^2 \le 4.$$ Now, let's substitute $x = b + c$, $y = c + a$, and $z = a + b$. The problem is now equivalent to showing that $$x^2 + y^2 + z^2 \le 4 \Longrightarrow \frac{\left(\frac{-x+y+z}{2} \cdot \frac{x-y+z}{2} + 1 \right)}{z^2} + \frac{\left(\frac{x-y+z}{2} \cdot \frac{x+y-z}{2} + 1 \right)}{y^2} + \frac{\left(\frac{x+y-z}{2} \cdot \frac{-x+y+z}{2} + 1 \right)}{x^2}.$$ We can rewrite $$f(x, y, z) = \frac{\left(\frac{-x+y+z}{2} \cdot \frac{x-y+z}{2} + 1 \right)}{z^2} + \frac{\left(\frac{x-y+z}{2} \cdot \frac{x+y-z}{2} + 1 \right)}{y^2} + \frac{\left(\frac{x+y-z}{2} \cdot \frac{-x+y+z}{2} + 1 \right)}{x^2} = \frac{z^2 - (x - y)^2 + 4}{4z^2} + \frac{y^2 - (x - z)^2 + 4}{4y^2} + \frac{x^2 - (z - y)^2 + 4}{4x^2},$$which we can rewrite as $$f(x, y, z) = \frac{z^2 - (x - y)^2 + 4}{4z^2} + \frac{y^2 - (x - z)^2 + 4}{4y^2} + \frac{x^2 - (z - y)^2 + 4}{4x^2} = \frac{3}{4} - \frac{x^2 + y^2}{4z^2} - \frac{x^2 + z^2}{4y^2} - \frac{y^2 + z^2}{4x^2} + \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} + \frac{zy}{2x^2} + \frac{zx}{2y^2} + \frac{xy}{2z^2}.$$We know that $x^2 + y^2 + z^2 \le 4$, so $-x^2 - y^2 \ge z^2 - 4$, etc, from which we have that \begin{align*} f(x, y, z) &\ge \frac{3}{4} + \frac{z^2 - 4}{4z^2} + \frac{y^2 - 4}{4y^2} + \frac{x^2 - 4}{4x^2} + \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} + \frac{zy}{2x^2} + \frac{xz}{2y^2} + \frac{xy}{2z^2} \\ &= \frac{3}{2} + \frac{zy}{2x^2} + \frac{xz}{2y^2} + \frac{xy}{2z^2}. \end{align*}By AM-GM, $\frac{zy}{2x^2} + \frac{xz}{2y^2} + \frac{xy}{2z^2} \ge \frac{3}{2}$, which gives $f(x, y, z) \ge 3$ as desired.

The given condition expands and simplifies to $a^2+b^2+c^2+ab+bc+ca\le 2$. We wish to show that $$\frac{2ab+2}{(a+b)^2}+\frac{2bc+2}{(b+c)^2}+\frac{2ca+2}{(c+a)^2}\ge 6$$. If we can show that $$\sum_{\text{cyc}} \frac{2ab+a^2+b^2+c^2+ab+bc+ca}{(a+b)^2}\ge 6$$we would be done. Note that $$\frac{2ab+a^2+b^2+c^2+ab+bc+ca}{(a+b)^2} = 1 + \frac{(c+a)(c+b)}{(a+b)^2}$$, so we want to show: $$\sum_{\text{cyc}} \frac{(c+a)(c+b)}{(a+b)^2}\ge 3$$. However, this is obviously true by AM-GM, so we're done. $\blacksquare$

we can rewrite the condition to be $2\ge a^2+b^2+c^2+ab+bc+ac$. now we have $$\sum\frac{2ab+2}{(a+b)^2}\ge\sum\frac{2ab+a^2+b^2+c^2+ab+bc+ac}{(a+b)^2}=\sum\frac{(a+b)^2+(c+a)(c+b)}{(a+b)^2}=$$$$3+\sum\frac{(c+a)(c+b)}{(a+b)^2}$$now by AM-GM, we are done

ugly sol \begin{align*} \sum_{cyc} \frac{ab+1}{(a+b)^2} &= \frac{1}{4} \sum_{cyc} \frac{4ab+4}{(a+b)^2} \\ &\ge \frac{1}{4} \sum_{cyc} \frac{4ab+\left( a^2 + b^2 + c^2 + (a+b+c)^2 \right)}{(a+b)^2} \\ &= \frac{1}{4} \sum_{cyc} \frac{2a^2 + 2b^2 + 2c^2 + 6ab + 2bc + 2ca}{(a+b)^2} \\ &= \frac{1}{4} \sum_{cyc} 2 + \frac{2c^2 + 2ab + 2bc + 2ca}{(a+b)^2} \\ &= \frac{3}{2} + \frac{3}{2} \cdot \frac{1}{3} \sum_{cyc} \frac{ (c+a)(c+b)}{(a+b)^2} \\ &\ge \frac{3}{2} + \frac{3}{2} \prod_{cyc} \frac{(c+a)(c+b)}{(a+b)^2} \\ &= 3. \end{align*}

We know that $a^2+b^2+c^2+ab+bc+ca\le 2$. Note that $$\sum_{cyc}\frac{2ab+2}{(a+b)^2}$$$$\ge \sum_{cyc}\frac{2ab+(a^2+b^2+c^2+ab+bc+ca)}{(a+b)^2}$$$$=\sum_{cyc}\frac{(a+b)^2+(c+a)(c+b)}{(a+b)^2}$$$$=3+\sum_{cyc}\frac{(c+a)(c+b)}{(a+b)^2}.$$By AM-GM, this is $$\ge 3 + 3\sqrt[3]{1} = 6.$$This completes the problem.

Solution Since $a^2+b^2+c^2+(a+b+c)^2\leq4 \implies 2(a^2+b^2+c^2+ ab+bc+ca)\leq4, $since $ ab+bc+ca \leq a^2+b^2+c^2 \implies ab+bc+ca \leq 1$ We also get get $a^2+b^2+c^2 \leq 2$ . By RMS-AM we have that $ \sqrt 6 \geq a+b+c$ By AM-HM we have $\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ca+1}{(c+a)^2}\geq \frac{9}{\frac{(a+b)^2}{ab+1}+\frac{(b+c)^2}{bc+1}+\frac{(c+a)^2}{ca+1}}$ Now we can prove that $3 \geq \frac{(a+b)^2}{ab+1}+\frac{(b+c)^2}{bc+1}+\frac{(c+a)^2}{ca+1}$ by applying Titu's 2 times and using the above results

$a^2+b^2+c^2+(a+b+c)^2\leq4$ = $2(a^2+b^2+c^2)+2(ab+bc+ac)\leq4$ = $a^2+b^2+c^2+ab+bc+ca\leq2$ Now, $\sum_{cyc} \frac{ab + 1}{(a+b)^2}$ = $\frac{1}{2} \sum_{cyc} \frac{2ab + 2}{(a+b)^2}$ $\geq$ $\frac{1}{2} \sum_{cyc} \frac{2ab + a^2+b^2+c^2+ab+bc+ca}{(a+b)^2}$ = $\frac{1}{2} \sum_{cyc} \frac{(a+b)^2+c^2+ab+bc+ca}{(a+b)^2}$ = $\frac{1}{2} \sum_{cyc} \frac{(a+b)^2+(a+c)(b+c)}{(a+b)^2}$ = $\frac{3}{2}+\frac{1}{2} \sum_{cyc} \frac{(a+c)(b+c)}{(a+b)^2}$ However, we know that $\frac{yz}{x^2} \frac{xz}{y^2} \frac{yx}{z^2} \geq 3$ Now by setting $x = (a+b)$ $y = (c+b)$ $z = (a+c)$ We get that $\sum_{cyc} \frac{(a+c)(b+c)}{(a+b)^2} \geq 3$ Therefore, $\frac{3}{2}+\frac{1}{2} \sum_{cyc} \frac{(a+c)(b+c)}{(a+b)^2} \geq \frac{3}{2} + \frac{3}{2}$ = $3$ $\mathbb{QED}$

The condition is equivalent to $2 \geq a^2+b^2+c^2+ab+bc+ca.$ Therefore, $$\text{LHS} = \sum_{cyc} \frac{2ab+2}{2(a+b)^2} \geq \sum_{cyc} \frac{a^2+b^2+2ab+c^2+ab+bc+ca}{2(a+b)^2} = \frac32 + \frac12 \sum_{cyc} \frac{(c+a)(c+b)}{(a+b)^2} \geq 3$$by AM-GM. QED