Let $a, b, c$ be positive real numbers such that $a^2+b^2+c^2+(a+b+c)^2\leq4$. Prove that \[\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ca+1}{(c+a)^2}\geq 3.\]
Problem
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Tags: USA(J)MO, USAMO, inequalities
29.04.2011 00:51
This was USAJMO #2, not 3.
29.04.2011 00:56
dang. that is pretty tricky numbering. anyways
29.04.2011 00:57
29.04.2011 01:00
29.04.2011 01:26
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?
29.04.2011 01:43
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!
29.04.2011 01:55
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.
29.04.2011 03:27
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 .
29.04.2011 03:32
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 $(ab+1)(bc+1)(ac+1)$.
29.04.2011 03:50
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...
29.04.2011 04:13
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?
29.04.2011 04:22
So I did a different substitution than most!
29.04.2011 04:45
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
29.04.2011 05:01
I used triangle notation, homogenized, multiplied everything out, then used two applications of Muirhead.
29.04.2011 05:38
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 ...
29.04.2011 06:01
Would we be docked if we didn't explicitly state when equality occurs?
29.04.2011 06:11
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.
29.04.2011 06:30
Does anyone know how to solve my "harder" version of the problem, by the way?
29.04.2011 06:36
I don't know if it is true. Plug in $a=b=c=\frac{1}{\sqrt{3}}$.
03.08.2023 03:20
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.
04.11.2023 10:34
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 $a=b=c=\frac{1}{\sqrt3}$
13.02.2024 08:11
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. $\Box$
23.02.2024 22:43
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.
16.03.2024 18:40
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, $\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ca+1}{(c+a)^2}\geq 3$
17.04.2024 06:43
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.
14.05.2024 00:13
$$a^2 + b^2 + c^2 + (a + b + c)^2 \le 4 \Longleftrightarrow a^2 + b^2 + c^2 + ab + bc + ca \le 2 \Longleftrightarrow (a + b)^2 + (b + c)^2 + (c + a)^2 \le 4.$$ 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.
07.07.2024 02:04
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. $\blacksquare$
24.08.2024 21:50
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
25.08.2024 08:01
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*}
04.11.2024 23:19
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.
25.12.2024 13:51
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
27.12.2024 16:59
$a^2+b^2+c^2+(a+b+c)^2\leq4$ = $2(a^2+b^2+c^2)+2(ab+bc+ac)\leq4$ = $a^2+b^2+c^2+ab+bc+ca\leq2$ 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, $\frac{3}{2}+\frac{1}{2} \sum_{cyc} \frac{(a+c)(b+c)}{(a+b)^2} \geq \frac{3}{2} + \frac{3}{2}$ = $3$ $\mathbb{QED}$
27.12.2024 20:42
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