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This is a high school contest exercise question.

If $a,b,c\ge0$ and $a^2+b^2+c^2=2$, show that $\frac{a^2}{a^2+2bc+1}+\frac{b^2}{b^2+2ca+1}+\frac{c^2}{c^2+2ab+1}\le1$.

It is straightforward to show that the quantity is $\ge \frac{2}{3}$, where equality holds when $a=b=c$ or when one of them is $2$ and the other two being zero. But my attempt to find the upper bound was less smooth.

My main question is what common strategies we should try when dealing with inequalities of which the equality sign holds on the boundary. My limited knowledge on inequalities usually suggests the equality sign holds when all variables are equal or are proportional to each other. But the quantity above has maximum when some variable is zero and others being equal.

Failed attempts:

  1. The only inequality that I know to have the above feature is Schur's inequality. But I could not figure out how to use Schur's inequality in this problem.
  2. I also tried to redefine $C=1-c$ and study the inequality in terms of $a$, $b$, and $C$, hoping that the familiar $a=b=C$ situation will show up. But I did not succeed in this path either.

My proof:

  1. Assume $c$ to be smallest among $a$, $b$, and $c$. Then each of the $bc$, $ca$, and $ab$ in the denominators might be replaced by $c^2$ to get a larger quantity.
  2. For fixed $c$, the resulting first two terms $\frac{a^2}{a^2+2c^2+1}$ and $\frac{b^2}{b^2+2c^2+1}$ are concave functions of $a^2$ and $b^2$. So, by using Jensen's inequality, their sum is bounded above by $$2\cdot\frac{\frac{2-c^2}{2}}{\frac{2-c^2}{2}+2c^2+1}.$$
  3. Now, together with the third term left out earlier, there is only a single variable $c^2$. I can check, by calculating the derivative, that the sum is decreasing over $c^2\ge 0$ so that the maximum $1$ occurs when $c^2=0$.

I don't quite like the above proof mostly because the problem is symmetric, but my proof treated variables differently. Is there a better proof?

PS: I'm more interested in inequality-based strategies. So KKT conditions / Lagrange multipliers are not what I am looking for currently.

Zack Fisher
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    if you have convex function $f$, then you have $f(a)+f(b)\geq 2f\left(\frac{a+b}{2}\right)$ – dezdichado Dec 31 '24 at 20:24
  • @dezdichado Thank you so much. That was indeed my typo. It should be concave. Just edited the post. – Zack Fisher Dec 31 '24 at 20:28
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    One common start when you got something that achieves equality on a case like $a=b$ and $c=0,$ is to see if you can prove something like: $$G(a,b,c)\leq G(m(a,b), m(a,b),c)$$ which turns your inequality into a single-variable one if the equality is homogeneous, where $m(a,b)$ is some "average" type function such as $\frac{a+b}{2}$ or $\sqrt{ab}$ and so on. – dezdichado Dec 31 '24 at 20:28
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    On the other hand, for a symmetric, homogeneous inequality like this the biggest hammer is the EV theorem.

    But usually, something milder like the $uvw$ method, some clever subs, some Jensen's, rearrangements and their combinations are good enough.

     – dezdichado Dec 31 '24 at 20:30
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    @dezdichado Thank you so much for the nice advice. I'll learn and try them. – Zack Fisher Dec 31 '24 at 20:35
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    I did not check the details of your solution, but if it's correct then it's a pretty nice one. Although the general rule I have in my head is that if something yields to rearrangement, then it is not a super tight inequality and there are likely other solutions. – dezdichado Dec 31 '24 at 20:42

3 Answers3

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We have $$\frac{a^2}{a^2+2bc+1} = \frac{a^2}{1+a^2}\cdot \frac{1}{1 + \frac{2bc}{1+a^2}} \le \frac{a^2}{2a}\cdot \frac{1}{1 + \frac{2bc}{1+(2-2bc)}} = \frac{a}{2} - \frac13 abc$$ where we use $1+a^2 \ge 2a$, and $a^2 = 2- b^2-c^2 \le 2 - 2bc$.

It suffices to prove that $$\frac{a+b+c}{2} - abc \le 1.$$

We have \begin{align*} a + b + c - 2abc &\le a + b + c - abc\\ &= 1\cdot (a + b) + (1-ab)\cdot c\\ &\le \sqrt{1^2 + (1-ab)^2}\sqrt{(a+b)^2 + c^2}\\ &= \sqrt{2-2ab + a^2b^2}\sqrt{2 + 2ab}\\ &\le 2 \end{align*} where we use the Cauchy-Bunyakovsky-Schwarz inequality, and $ab \le \frac{a^2 + b^2}{2} \le 1$, and $4 - (2-2ab + a^2b^2) (2 + 2ab) = 2a^2b^2(1-ab)\ge 0$.

We are done.

River Li
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  • Thank you so much for the beautiful proof. Would you please explain your thought process for this problem? (1) For the initial inequality, what motivates you to rewrite the terms that way? (2) For the $a+b+c-2abc\le a+b+c-abc$, did you already have the idea that $abc=0$ is needed for the maximum to be reached? If so, why didn't you try $a+b+c-2abc\le a+b+c$ instead? – Zack Fisher Jan 01 '25 at 21:59
  • @ZackFisher (1) Experience; Try and error. (2) First, we guess some cases $a=b=c$, or $abc=0$, or $a=b$ for many inequalities. Second, we have $a + b + c - 2abc = a + b + (1 - 2ab)c$. Because $ab \le 1$, $1-2ab$ can be negative, so I try $a+b+(1-ab)c$. Yes, I also try $a + b + c \le 2$ but it is not true (check the case $a = b$). – River Li Jan 02 '25 at 00:57
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    Thanks for the explanation! – Zack Fisher Jan 02 '25 at 03:29
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Just got another fabulous proof from Mick Fang.

$$a(b+c)\le \frac{a^2 +(b+c)^2}2=1+bc\le 1+2bc.$$ Thus, $$a(a+b+c) \le a^2+2bc+1, $$i.e., $$\frac{a^2}{a^2+2bc+1}\le \frac{a}{a+b+c}.$$ Similarly, $$\frac{b^2}{b^2+2ca+1}\le \frac{b}{a+b+c},$$ and $$\frac{c^2}{c^2+2ab+1}\le \frac{c}{a+b+c}.$$ Adding the above three inequalities completes the proof.

Zack Fisher
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First, homogenize by substituting in your expression $S$ in the denominator $1$ with $(a^2+b^2+c^2)/2$. Now, in the difference $1-S$ do as usual $a=x$, $b=x+y$, $c=x+y+z$, and get in the numerator an expression with all monomials with positive coefficients. We get equality if and only if $x=z=0$, that is $a=0$ and $b=c (=1)$. This is all quite standard.

$\bf{Added:}$

We have $$1-S=(a + b + c)^2 (a^4 + 2 a^3 b - 6 a^2 b^2 + 2 a b^3 + b^4 + 2 a^3 c + 10 a^2 b c + 10 a b^2 c + 2 b^3 c - 6 a^2 c^2 + 10 a b c^2 - 6 b^2 c^2 + 2 a c^3 + 2 b c^3 + c^4)/ / (a^2 + 3 b^2 + 4 a c + c^2) (3 a^2 + b^2 + 4 b c + c^2) (a^2 + 4 a b + b^2 + 3 c^2)$$

Now you have to show that the symmetric expression in $a$, $b$, $c$ $$(a^4 + 2 a^3 b - 6 a^2 b^2 + 2 a b^3 + b^4 + 2 a^3 c + 10 a^2 b c + 10 a b^2 c + 2 b^3 c - 6 a^2 c^2 + 10 a b c^2 - 6 b^2 c^2 + 2 a c^3 + 2 b c^3 + c^4)\ge 0$$ for all $a$,$b$,$c\ge 0$. Note that we do not have equality for all the numbers equal, however, we have equality if two are equal, and the third $0$. In any case, we can assume $0\le a\le b\le c$, so write $a=x$, $b=x+y$, $c=x+y+z$, with $x$, $y$, $z\ge 0$ and we get the expression

$$27 x^4 + 72 x^3 y + 70 x^2 y^2 + 24 x y^3 + 36 x^3 z + 70 x^2 y z + 36 x y^2 z + 16 x^2 z^2 + 28 x y z^2 + 6 y^2 z^2 + 8 x z^3 + 6 y z^3 + z^4$$ which is $\ge 0$ for $x$, $y$, $z\ge 0$, with equality if ( and only if ) $x=z=0$, that is $a=0$, and $b=c$ ( $= 1$ with the extra condition $a^2+b^2+c^2=2$).

Note: the method works for some inequalities, like in this case

orangeskid
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