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Let $a,b,c$ be non-negative real numbers such that $a^2+b^2+c^2=3$. Find the minimum value of $$P=ab+bc+3ca+\dfrac{3}{a+b+c}.$$ This is an asymmetric inequality. It is hard for me to find when the equation holds. I guess when it occurs if $a=c=0$ and $b=\sqrt{3}$. Then $\min P=\sqrt{3}$. But I have no idea to solve it. Please help me a hint. Thank you.

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    $\begingroup$ Have you learned about Lagrange multipliers? $\endgroup$
    – user145413
    Commented Apr 3, 2021 at 10:38
  • $\begingroup$ A simiilar problem: math.stackexchange.com/questions/514695/… $\endgroup$
    – V.G
    Commented Apr 3, 2021 at 10:56

2 Answers 2

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As $a,c \ge 0$ then $$P \ge S := ab+bc+ca + \frac{3}{a+b+c}$$

We minimize $S$

As $a^2+b^2 +c^2 = 3$ then $$S =\frac{1}{2} (a+b+c)^2+\frac{3}{a+b+c} -\frac{3}{2}$$ Denote $x = a+b+c$. Because $a^2+b^2 +c^2 = 3$ then $\sqrt{3} \le x \le 3$ $$S = f(x) := \frac{1}{2}x^2+\frac{3}{x}-\frac{3}{2}$$ As $f'(x) = x-\frac{3}{x^2}>0$ in the interval $[\sqrt{3},3]$ then $f(x)$ is increasing. Hence, $S_1 = f(x)$ is minimized as $a+b+c = x_0 = \sqrt{3}$

Conclusion: we can conclude that $$P \ge S \ge \sqrt{3} $$ The equality occurs when $(a,b,c) = (0,\sqrt{3},0),(0,0,\sqrt{3})$ or $(\sqrt{3},0,0)$

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  • $\begingroup$ Thank you very much. Your solution is simpler than what I think it has to. $\endgroup$
    – Chivul
    Commented Apr 3, 2021 at 17:46
  • $\begingroup$ @VichLee You're welcome! $\endgroup$
    – NN2
    Commented Apr 3, 2021 at 17:46
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No matter what, by simply calculating one has $P(\sqrt 3, 0, 0) = \sqrt 3$. On the other hand,

We know a minimum must exist by Weierstrass's theorem. This is because $Z := \{(a, b, c) \in \mathbb R ^3\mid a, b, c \geq 0, a^2 + b^2 + c^2 = 3\}$ is a closed bounded set and $P$ is continuous on $Z$.

Write $$P(a,b,c) = S_1 + S_2 \tag{1}$$

where $S_1 :=ab + bc + \dfrac{3}{2} + ac + \dfrac{3}{a + b + c}$ and $S_2 :=2ac - \dfrac{3}{2}$. Note that $S_1$ is symmetric. The choice of $\dfrac{3}{2}$ will be clear should you read on.

If it is possible to simultaneously minimise $S_1$ and $S_2$ (with given constraints), then their sum will be the minimum value of $P$. Let us try to minimise $S_1$.

Plugging $bc = \dfrac{1}{2}(b+c)^2 - \dfrac{3}{2} + \dfrac{a^2}{2}$ into $S_1$ yields $$ S_1 = \dfrac{1}{2}(b+c)^2 + (b + c) a + \frac{3}{(b + c) + a} + \dfrac{a^2}{2},$$

or with $x := b + c$, \begin{align} \begin{split} S_1(x, a) &= \dfrac{1}{2} x^2 + a\, x + \frac{3}{x + a} + \dfrac{a^2}{2} = \\ &= \dfrac{1}{2} (x + a)^2 + \dfrac{3}{x + a} = \\ &= \dfrac{1}{2} z^2 + \dfrac{3}{z}. \end{split} \tag{2} \end{align}

  • We are left with the problem of minimising $S_1(z)$ wrt $z$ when $z \geq \sqrt 3$ (because $z^2 \geq a^2 + b^2 + c^2$).

  • The derivative with respect to $z$ is $z - \dfrac{3}{z^2}$. This is positive if $z > \sqrt[3]{3}$. So $S_1$ is increasing for $z \geq \sqrt[3]{3}$ and we should pick $z_0 := \sqrt 3$ if we can (choosing $z_0 := \sqrt[3]{3}$ is not possible because $\sqrt[3]{3} < \sqrt 3$).

If one moves up the chain, we have $z_0 = x_0 + a_0 = a_0 + b_0 + c_0 = \sqrt 3$. And we see it is possible to choose $a_0 := \sqrt 3$ and $b_0 := c_0 := 0$ to obtain $z_0 = \sqrt 3$, thus minimising $S_1$. By picking $c_0 = 0$, $S_2$ is minimised as well, so we are done. In conclusion, the minimum value of $P$, subject to given constraints, is $\sqrt 3$. It is obtained at, for example, $(a_0, b_0, c_0) = (\sqrt 3, 0, 0)$.

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  • $\begingroup$ No, the second term is minimized when $a=b=c=1$ $\endgroup$
    – NN2
    Commented Apr 3, 2021 at 11:31
  • $\begingroup$ @NN2 Hopefully, the answer has now improved. $\endgroup$ Commented Apr 3, 2021 at 13:32
  • $\begingroup$ I read until this equation and I stop because this equation $$ (b + c) + (b + c)\alpha + \frac{3}{(b + c) + 1 + \alpha}$$ is false. You forgot at least the factor $3$ of $3ac$ $\endgroup$
    – NN2
    Commented Apr 3, 2021 at 15:19
  • $\begingroup$ @NN2 I partitioned $P$ into $S_1$ and $S_2$ to get rid of said factor. $\endgroup$ Commented Apr 3, 2021 at 15:21
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    $\begingroup$ @NN2 I fixed my answer after a good night's sleep. But I see in the meantime you uploaded your answer (+1) which I think is more clear. $\endgroup$ Commented Apr 4, 2021 at 12:38

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