Find the minimum value of $P=ab+bc+3ca+\dfrac{3}{a+b+c}$

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.

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

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 $$(\sqrt{3},0,0)$$

• Thank you very much. Your solution is simpler than what I think it has to. Commented Apr 3, 2021 at 17:46
• @VichLee You're welcome!
– NN2
Commented Apr 3, 2021 at 17:46

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, $$(a_0, b_0, c_0) = (\sqrt 3, 0, 0)$$.

• No, the second term is minimized when $a=b=c=1$
– NN2
Commented Apr 3, 2021 at 11:31
• @NN2 Hopefully, the answer has now improved. Commented Apr 3, 2021 at 13:32
• 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 $3ac$
– NN2
Commented Apr 3, 2021 at 15:19
• @NN2 I partitioned $P$ into $S_1$ and $S_2$ to get rid of said factor. Commented Apr 3, 2021 at 15:21
• @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. Commented Apr 4, 2021 at 12:38