$a,b,c$ are positive real numbers such that $ab+bc+ca=1$. What is the minimum value of $10a^2+10b^2+c^2$?

I want to solve this problem without using Lagrange multipliers or calculus. I tried the following with some basic inequalities:

From AM-GM inequality $$5a^2+5b^2\geq10ab\\ 5b^2+\frac 12c^2\geq\sqrt{10}bc\\ \frac 12c^2+5a^2\geq\sqrt{10}ca$$ Summing them gives $$10a^2+10b^2+c^2\geq\sqrt{10}(\sqrt{10}ab+bc+ca)$$ But this doesn't help.

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    $\begingroup$ Thats a good idea, but try different coefficients. In particular, what is the equality case likely to be? $\endgroup$
    – Calvin Lin
    Sep 7, 2021 at 5:58
  • $\begingroup$ I don't understand the point of not using Lagrange multipliers when that's the obvious strategy. Are we trying to find some insight into Lagrange multipliers here that would otherwise be obfuscated by procedure? $\endgroup$
    – Jared
    Sep 7, 2021 at 6:24
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    $\begingroup$ @Jared Because it's a contest-math question, where calculus isn't needed and students are encouraged/suggested to find a non-calculus solution. $\endgroup$
    – Calvin Lin
    Sep 7, 2021 at 6:25
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    $\begingroup$ @Jared I'm commenting based on the context OP has provided (and my assumptions). I agree that Lagrange multipliers is a straightforward approach for those who know it well. $\quad $ As another consideration, what if you had to explain this to a 8 year old? Would you require them to learn about calculus and LM first, or show them an easier AM-GM approach? Of course, you may never care about such a scenario, just like you're not interested in contests. $\endgroup$
    – Calvin Lin
    Sep 7, 2021 at 6:33
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    $\begingroup$ @Jared I don't want a Lagrange multiplier solution because I don't know it as I'm a high schooler. Also as I am the asker of the question, I think I have the right to mention what kind of answer I need so that the answer is suitable for me. $\endgroup$
    – Oshawott
    Sep 7, 2021 at 6:43

4 Answers 4


Hint: If we additionally make the (reasonable, though not necessarily true) assumption that the minimum happens when $ a = b$, then reduce it to a 1 variable inequality and hence show that the minimum is 4 obtained at $ ( 1/3, 1/3, 4/3)$.
Note that we've not proven this is the minimum yet.

Use this to figure out what AM-GM's to create.
Modifying your approach, we have the following inequalities:

$a^2 + b^2 \geq 2ab \longrightarrow$ This reinforces $a=b$ at the equality case.
$Xa^2 + 0.5c^2 \geq 2\sqrt{X/2}ac \longrightarrow$ What does the (assumed) equality case tell us about $X$?
$Yb^2 + 0.5c^2 \geq 2\sqrt{Y/2}bc \longrightarrow$ What does the (assumed) equality case tell us about $Y$?

From the equality case, we require

$ X = 8$, $ Y = 8$.

Then, weight the inequalities to get

$ 2a^2 + 2b^2 \geq 4ab $
$ 8 a^2 + 0.5 c^2 \geq 4 ac $
$ 8b^2 + 0.5c^2 \geq 4bc $
Summing them up gives:

$$ 10a^2 + 10b^2 + c^2 \geq 4(ab+bc+ca) = 4,$$

with equality when $ a = b = c/4, ab+bc+ca = 1$, IE $ ( 1/3, 1/3, 4/3)$.
(Thankfully, our initial assumption is true.)

I wanted to show you how to derive these inequalities (assuming that your stated approach could work). It isn't just "magic" or "by observation" or "by luck".
For the actual solution, you just need to write out the 3 AM-GM inequalities, sum them up, and verify the equality case.

Using this approach, try your hand at minimizing (say) $3a^2 + 2b^2 + c^2 $ given $ab+bc+ ca = 1$.
(Note: I've not actually done this, so I can't guarantee that the values will look nice.)

  • $\begingroup$ There is reasoning, how to find the minimum without assumption that the equality occurs for $a=b$. See my solution. This reasoning works also in the general. $\endgroup$ Sep 7, 2021 at 6:56
  • $\begingroup$ @MichaelRozenberg Right. I wanted to stay within OP's approach (and without adding too many unknowns, hence the assumption $a=b$). EG My go-to for quadratic inequalities is to set up the matrix and calculate discriminants. $\endgroup$
    – Calvin Lin
    Sep 7, 2021 at 7:01
  • $\begingroup$ @CalvinLin If AM-GM (actually) works, then we can do it using the method of unknown coefficients: $p_1 a^2 + q_1b^2 \ge 2\sqrt{p_1q_1}ab, \quad p_2b^2 + q_2c^2 \ge 2\sqrt{p_2q_2}bc,\quad p_3c^2 + q_3a^2 \ge 2\sqrt{p_3q_3}ca$, and we require $p_1q_1 = p_2q_2 = p_3q_3$, $p_1 + q_3 = 10, q_1 + p_2 = 10, q_2 + p_3 = 1$. $\endgroup$
    – River Li
    Sep 7, 2021 at 7:06
  • $\begingroup$ @RiverLi Right, hence the "without adding too many unknowns", esp since I'm assuming that OP isn't familiar with it. What I'm doing is essentially a simplified version. $\endgroup$
    – Calvin Lin
    Sep 7, 2021 at 7:09
  • $\begingroup$ @CalvinLin Sometimes we might do it in such a way, for example, the following problem (and solution due to SBM@AoPS): Let $a, b \ge 0$. Prove that $a^2b^2 (a^2 + b^2 - 2) \ge (a + b)(ab - 1)$. We can split into three parts (AM-GM): ${a}^{2}b\le \frac25\,{a}^{4}{b}^{2}+\frac25\,a+\frac15\,b$, $a{b}^{2}\le\frac{1}{10}\,{a}^{4}{b}^{2}+\frac{3}{10}\,{a}^{2}{b}^{4}+\frac{3}{5}\,b$, and $2\,{a}^{2}{b}^{2}\le \frac12{a}^{4}{b}^{2}+{\dfrac {7}{10}}{a}^{2}{b}^ {4}+\frac35a+\frac15b$. $\endgroup$
    – River Li
    Sep 7, 2021 at 7:20

Let $k$ be the minimal value.

Thus, $k>0$ and the inequality $$10a^2+10b^2+c^2\geq k(ab+ac+bc)$$ or $$c^2-k(a+b)c+10a^2+10b^2-kab\geq0$$ is true for any reals $a$,$b$ and $c$ (because after homogenization the condition $ab+ac+bc=1$ is not relevant already), which says $$k^2(a+b)^2-4(10a^2+10b^2-kab)\leq0,$$ which gives $$(40-k^2)a^2-2(2k+k^2)ab+(40-k^2)b^2\geq0,$$ for which we need $40-k^2>0$ and $$(2k+k^2)^2-(40-k^2)^2\leq0$$ or $$(k+20)(k+5)(k-4)\leq0$$ or $$0<k\leq4,$$ which gives that a maximal value of $k$, for which the inequality $$10a^2+10b^2+c^2\geq k(ab+ac+bc)$$ is true it's $4$.

The equality occurs for $a=b$ and $c=\frac{k(a+b)}{2}=4a$.

Id est, $$\min_{ab+ac+bc=1}(10a^2+10b^2+c^2)=4.$$

  • 1
    $\begingroup$ Side note: You might want to clarify why it's "true for any reals $a, b, c$", since the question is restricted to $ ab + bc + ca = 1$. $\endgroup$
    – Calvin Lin
    Sep 7, 2021 at 7:15
  • $\begingroup$ @Calvin Lin I added. Thank you! $\endgroup$ Sep 7, 2021 at 7:36
  • $\begingroup$ typo: $(40-k^2)a^2-2(2k-k^2)ab+(40-k^2)b^2\geq0$ should be $(40-k^2)a^2-2(2k+k^2)ab+(40-k^2)b^2\geq0$. $\endgroup$
    – River Li
    Sep 7, 2021 at 9:38
  • $\begingroup$ @River Li Thank you! I fixed. $\endgroup$ Sep 7, 2021 at 9:55

Noting that $$20 a^{2}+20 b^{2}+2 c^{2}-8(a b+b c+c a)=(2 a-2 b)^{2}+(4 b-c)^{2}+(4 a-c)^{2} \geqslant 0, $$ we have $$ \begin{aligned} \quad 2\left(10 a^{2}+10 b^{2}+c^{2}\right)\geqslant 8 \quad \Rightarrow \quad 10 a^{2}+10 b^{2}+c^{2} \geqslant 4 \end{aligned} $$ The minimum value of $10 a^{2}+10 b^{2}+c^{2}$ is $4$, which is attained if and only if$$ \left\{\begin{array}{l} a=b \\ 4 b=c \\ c=4 a \end{array}\right. \Leftrightarrow a=b=\frac{1}{3} \text { and } c=\frac{4}{3} $$


Another way.

By C-S $$10a^2+10b^2+c^2=$$ $$=\sqrt{\left(8a^2+\frac{c^2}{2}+8b^2+\frac{c^2}{2}+2a^2+2b^2\right)\left(\frac{c^2}{2}+8a^2+\frac{c^2}{2}+8b^2+2b^2+2a^2\right)}\geq\sqrt{(2ac+2ac+2bc+2bc+2ab+2ab)^2}=\sqrt{16}=4.$$ The equality occurs, when $$\left(2\sqrt2a,\frac{c}{\sqrt2},2\sqrt2b,\frac{c}{\sqrt2},\sqrt2a,\sqrt2b\right)||\left(\frac{c}{\sqrt2},2\sqrt2a,\frac{c}{\sqrt2},2\sqrt2b,\sqrt2b,\sqrt2a\right),$$ which gives $a=b=\frac{c}{4}$, which says that we got a minimal value.


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