Finding $\small{\min\limits_{a+b+c=2}\sqrt{a-2bc+3}+\sqrt{b-2ca+3}+\sqrt{c-2ab+3}.}$

Question

Given non-negative real numbers $a,b,c$ satisfying $a+b+c=2.$ Find the minimal value of expression $$P=\sqrt{a-2bc+3}+\sqrt{b-2ca+3}+\sqrt{c-2ab+3}.$$ Source: AOPS-Jokehim. See also here.

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Here is what I've done so far.

By $a=b=1;c=0$ I get $P=5$ so we need to prove that$$\sqrt{a-2bc+3}+\sqrt{b-2ca+3}+\sqrt{c-2ab+3}\ge 5. $$ I thought of Lagrange multiplier method.

We can split inequality into two cases.

$\bullet$ If $a=0;b+c=2,$ it's $\sqrt{3-2bc}+\sqrt{b+3}+\sqrt{c+3}\ge 5,$which is easy to check $c=2-b$ $$\sqrt{3-2b(2-b)}+\sqrt{b+3}+\sqrt{5-b}\ge 5,$$ where $0\le b\le 2.$

$\bullet$ $abc\neq0,$ we consider$$f(a,b,c,\lambda)=\sqrt{a-2bc+3}+\sqrt{b-2ca+3}+\sqrt{c-2ab+3}-5+\lambda(a+b+c-2).$$ and $(a,b,c)$ be an inside minimum point of $f$.

Thus, $$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=0,$$ which gives \begin{align*} \frac{1}{2\sqrt{a-2bc+3}}-\frac{2c}{2\sqrt{b-2ca+3}}-\frac{2b}{2\sqrt{c-2ab+3}}+1&=\frac{1}{2\sqrt{b-2ca+3}}-\frac{2c}{2\sqrt{a-2bc+3}}-\frac{2a}{2\sqrt{c-2ab+3}}+1\\&=\frac{1}{2\sqrt{c-2ba+3}}-\frac{2a}{2\sqrt{b-2ca+3}}-\frac{2b}{2\sqrt{a-2bc+3}}+1. \end{align*} (Notice that $a-2bc+3,b-2ca+3,c-2ab+3>0$ since $bc\le\dfrac{(b+c)^2}{4}\le\dfrac{(a+b+c)^2}{4}=1.$)

Now, let in this point $a\neq b$ and $a\neq c$ and we need to prove $b=c.$

I stopped here. Hope you help me. Also, any ideas and comments are welcome.

Answer 1

Let $c=\min\{a,b,c\}$.

We'll prove that $$\sqrt{a-2bc+3}+\sqrt{b-2ac+3}\geq\sqrt{2(a+b)-4c(a+b)+12-\frac{1}{2}(a-b)^2}.$$ Indeed, we need to prove that: $$2\sqrt{(a-2bc+3)(a-2ac+3)}\geq a+b-2c(a+b)+6-\frac{1}{2}(a-b)^2$$ or $$\sqrt{(2a(a+b+c)-8bc+3(a+b+c)^2)(2b(a+b+c)-8ac+3(a+b+c)^2)}\geq$$ $$\geq(a+b)(a+b+c)-4c(a+b)+3(a+b+c)^2-(a-b)^2$$ or $$(a-b)^2(6a^2+16ab+6b^2-4ac-4bc-19c^2)\geq0,$$ which is obvious.

Id est, it's enough to prove that: $$\sqrt{2(a+b)-4c(a+b)+12-\frac{1}{2}(a-b)^2}+\sqrt{c-2ab+3}\geq5$$ or $$2\sqrt{(a+b)(a+b+c)-4c(a+b)+3(a+b+c)^2-\frac{1}{2}(a-b)^2}+\sqrt{2c(a+b+c)-8ab+3(a+b+c)^2}\geq5(a+b+c)$$ or $$4\left((a+b)(a+b+c)-4c(a+b)+3(a+b+c)^2-\frac{1}{2}(a-b)^2\right)(2c(a+b+c)-8ab+3(a+b+c)^2)\geq(4(a^2+b^2+c^2)+8ab+15c(a+b))^2.$$ Now, let $a=c+u$ and $b=c+v$, $u$ and $v$ are non-negatives, and we obtain: $$3(67u^2-66uv+67v^2)c^2+$$ $$+4(u+v)(37u^2-54uv+37v^2)c+2(u-v)^2(13u^2+34uv+13v^2)\geq0$$ and we are done!

Answer 2

Alternative proof (sketch)

We may use the isolated fudging.

After homogenization, it suffices to prove that, for all $a, b, c \ge 0$ with $a + b + c > 0$, $$\sum_{\mathrm{cyc}} \frac{\sqrt{aQ - 2bc + 3Q^2}}{Q} \ge 5$$ where $Q := (a + b + c)/2$.

It suffices to prove that, for all $a, b, c \ge 0$ with $a + b + c > 0$, $$\frac{\sqrt{aQ - 2bc + 3Q^2}}{Q} \ge 5 \cdot \frac{23a^2 + 73a(b + c) + 16(b^2 + c^2) + 24bc}{55(a^2 + b^2 + c^2) +170(ab + bc + ca)}. \tag{1}$$

If $a = 0$, it is easy.

If $a > 0$, WLOG, assume that $a = 1$. Let $p = b + c, q = bc$. Then $p^2 \ge 4q$.

(1) is written as $$f(q) := \frac{\sqrt{(1 + p)/2 - 2q + 3(1 + p)^2/4}}{(1 + p)/2} + \frac23 - \frac{70p^2 + 287p + 91}{33p^2 + 102p + 36q + 33} \ge 0.\tag{2}$$

CLearly, $f(q)$ is concave on $q \ge 0$. It is not difficult to prove that $f(0) \ge 0$ and $f(p^2/4)\ge 0$ for all $p \ge 0$. Thus, $f(q) \ge 0$ for all $q \in [0, p^2/4]$.

We are done.

Answer 3

Here is my second proof.

Remark: I used this trick several years ago for the problem: What is the smallest constant $p>0$ such that $$\sqrt{x + \frac{(y-z)^2}{p}} + \sqrt{y + \frac{(z-x)^2}{p}} + \sqrt{z+\frac{(x-y)^2}{p}}\le \sqrt{3}$$ holds for all non-negative real number $x, y, z$ satisfying $x+y+z=1$?

$\phantom{2}$

WLOG, assume that $a \ge b \ge c$.

Using $2bc = \frac{(b + c)^2 - (b - c)^2}{2} = \frac{(2 - a)^2 - (b - c)^2}{2}$, the desired inequality is equivalently written as $$\sum_{\mathrm{cyc}} \sqrt{12a + 4 - 2a^2 + 2(b - c)^2} \ge 10. \tag{1}$$

We can prove that \begin{align*} &\sqrt{12a + 4 - 2a^2 + 2(b - c)^2} + \sqrt{12b + 4 - 2b^2 + 2(c - a)^2}\\[6pt] &\qquad + \sqrt{12c + 4 - 2c^2 + 2(a - b)^2}\\[6pt] \ge{}& 2\sqrt{12\cdot \frac{a + b}{2} + 4 - 2\cdot \left(\frac{a + b}{2}\right)^2 + 2\left(\frac{a + b}{2} - c\right)^2} + \sqrt{12c + 4 - 2c^2}. \tag{2} \end{align*} (The proof of (2) is given at the end. )

From (1) and (2), it suffices to prove that $$2\sqrt{12\cdot \frac{a + b}{2} + 4 - 2\cdot \left(\frac{a + b}{2}\right)^2 + 2\left(\frac{a + b}{2} - c\right)^2} + \sqrt{12c + 4 - 2c^2} \ge 10$$ or (using $a + b = 2 - c$) $$2\sqrt{4c^2 - 10c + 16} + \sqrt{12c + 4 - 2c^2} \ge 10 \tag{3}$$ which is true for all $0 \le c \le 2$ (not difficult).

We are done.

$\phantom{2}$

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Proof of (2).

Let \begin{align*} x &:= \sqrt{12a + 4 - 2a^2 + 2(b - c)^2}, \\ y &:= \sqrt{12b + 4 - 2b^2 + 2(c - a)^2},\\ z &:= \sqrt{12c + 4 - 2c^2 + 2(a - b)^2},\\ u &:= \sqrt{12\cdot \frac{a + b}{2} + 4 - 2\cdot \left(\frac{a + b}{2}\right)^2 + 2\left(\frac{a + b}{2} - c\right)^2},\\ v &:= \sqrt{12c + 4 - 2c^2}. \end{align*} We have $x^2 - u^2 = u^2 - y^2 = 2(a - b)(3 - a - b + c) \ge 0$, and $y^2 - z^2 = 4(b - c)(3 + a - b - c) \ge 0$.

It suffices to prove that $$x + y + z \ge 2u + v$$ or $$(x - u) + (z - v)\ge (u - y)$$ or $$\frac{x^2 - u^2}{x + u} + \frac{z^2 - v^2}{z + v} \ge \frac{u^2 - y^2}{u + y}$$ or $$\frac{2(a - b)(3 - a - b + c)}{x + u} + \frac{2(a - b)^2}{z + v} \ge \frac{2(a - b)(3 - a - b + c)}{u + y}$$ or (using $x \ge u \ge y \ge z$) $$\frac{2(a - b)(3 - a - b + c)}{2x} + \frac{2(a - b)^2}{2z} \ge \frac{2(a - b)(3 - a - b + c)}{2y}$$ or $$\frac{3 - a - b + c}{x} + \frac{a - b}{z} \ge \frac{3 - a - b + c}{y}$$ or (using $y \ge z$) $$\frac{3 - a - b + c}{x} + \frac{a - b}{y} \ge \frac{3 - a - b + c}{y}$$ or $$\frac{3 - a - b + c}{x}\ge \frac{3 - 2a + c}{y}$$ or $$\frac{3 - a - b + c}{2xy}\ge \frac{3 - 2a + c}{2y^2}$$ or (by AM-GM) $$\frac{3 - a - b + c}{x^2 + y^2}\ge \frac{3 - 2a + c}{2y^2}$$ or (equivalently) $$-2(a + b)^2 + 17(a + b) - 19 \ge 0$$ which is true (using $4/3 \le a + b \le 2$).

We are done.