v_Enhance wrote: Positive reals $a$, $b$, and $c$ obey $\frac{a^2+b^2+c^2}{ab+bc+ca} = \frac{ab+bc+ca+1}{2}$. Prove that \[ \sqrt{a^2+b^2+c^2} \le 1 + \frac{\lvert a-b \rvert + \lvert b-c \rvert + \lvert c-a \rvert}{2}. \]Proposed by Evan Chen Let $ab+ac+bc=x$. Hence, $a^2+b^2+c^2=\frac{x^2+x}{2}$ and $x^2-x\geq0$. $\lvert a-b \rvert + \lvert b-c \rvert + \lvert c-a \rvert=\sqrt{\sum_{cyc}(a-b)^2+2\sum_{cyc}|(a-b)(a-c)|}\geq$ $\geq\sqrt{\sum_{cyc}(a-b)^2+2\sum_{cyc}(a-b)(a-c)}=\sqrt{4\sum_{cyc}(a^2-ab)}=\sqrt{2x^2-2x}$. Hence, it remains to prove that $\sqrt{x^2-x}+\sqrt2\geq\sqrt{x^2+x}$. But $\sqrt{x^2-x}+\sqrt2\geq\sqrt{x^2+x}\Leftrightarrow\sqrt{x^2-x}+\sqrt{x^2+x}\geq\sqrt2x$, which is true because $\sqrt{x^2-x}+\sqrt{x^2+x}\geq\sqrt{x^2-x+x^2+x}=\sqrt2|x|\geq\sqrt2x$.

We let $a+b+c = u$ and $ab+bc+ca = v$. The condition gives us that $\frac{u^2-2v}{v}=\frac{v+1}{2}$, so $2u^2=v^2+5v$. First, I prove that $v \ge 1$. This follows from that $2u^2=v^2+5v \ge 6v \implies v(v-1) \ge 0$. Now, $|a-b|+|b-c|+|c-a|$ $= \sqrt{(a-b)^2+(b-c)^2+(c-a)^2+2|a-b||b-c|+2|b-c||c-a|+2|c-a||a-b|}$ $\ge \sqrt{(a-b)^2+(b-c)^2+(c-a)^2+2(a-b)(a-c)+2(b-a)(b-c)+2(c-a)(c-b)} = \sqrt{4u^2-12v}$. It now suffices to prove that $\sqrt{u^2-2v} \le 1+\sqrt{u^2-3v}$. Squaring, we have to prove $u^2-2v \le 1+u^2-3v+2\sqrt{u^2-3v}$. Squaring, we have to prove $v^2-2v+1 \le 4u^2-12v=2v^2-2v$, i.e. $v^2 \ge 1$, which we proved already. $\blacksquare$

how does arqady get $\sqrt{x^2-x}+\sqrt2\geq\sqrt{x^2+x}\Leftrightarrow\sqrt{x^2-x}+\sqrt{x^2+x}\geq\sqrt2x$?

Plops wrote: how does arqady get $\sqrt{x^2-x}+\sqrt2\geq\sqrt{x^2+x}\Leftrightarrow\sqrt{x^2-x}+\sqrt{x^2+x}\geq\sqrt2x$? because $$(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b$$

WLOG $a \ge b \ge c$, let $a^2 + b^2 + c^2 = x, ab + bc + ca = y$. Then $2x = y(y+1)$. Thus $x-y = \frac{y(y-1)}{2}$. As $x \ge y$, so $\boxed{y \ge 1}$. Now $2(a-c)^2 \ge (a-b)^2 + ( b-c)^2 + (a-c)^2 = 2(x-y) = y(y-1)$, implying $\boxed{\frac{(a-c)^2}{x} \ge \frac{y-1}{y+1}}$. Dividing both sides of the original inequality by $\sqrt{x}$, we obtain it suffices to show: \begin{align*} & \qquad ~~ 1 \le \frac{1}{\sqrt{x}} + \sqrt{\frac{y-1}{y+1}} \\ &\iff 1 \le \frac{1}{x} + \frac{y-1}{y+1} + 2 \sqrt{\frac{y-1}{x(y+1)}} \\ &\iff \frac{2}{y+1} \le \frac{1}{x} + 2 \sqrt{\frac{y-1}{x(y+1)}} \\ &\iff 1 \le \frac{y+1}{2x} + \sqrt{\frac{(y-1)(y+1)}{x}} \\ & \iff 1 \le \frac{1}{y} + \sqrt{ \frac{2(y-1)}{y} } \\ & \iff \frac{y-1}{y} \le \sqrt{2} \cdot \sqrt{\frac{y-1}{y}} \end{align*}As $\frac{y-1}{y} \le 1$ and $\sqrt{2} \ge 1$, so we are done! $\blacksquare$

Assume WLOG that $a \ge b \ge c$. Let $x=a^2+b^2+c^2$ and $y=ab+bc+ca$. The condition converts to $x=\tfrac{y(y+1)}{2}$, and we want to prove that $a-c \ge \sqrt{x}-1$. We also have \[2(x-y)=(a-b)^2+(b-c)^2+(c-a)^2 \le 2(a-c)^2,\]so $a-c \ge \sqrt{x-y}$. If $y \le 1$, then we have $\sqrt{x}-1 \le 0$, so $a-c \ge \sqrt{x}-1$. Otherwise, we have \begin{align*} 2y^2+2y \ge y^2+2y+1 &\implies 4x \ge (y+1)^2 \\ &\implies 2\sqrt{x} \ge y+1 \\ &\implies x-y \ge x-2\sqrt{x}+1 \\ &\implies \sqrt{x-y} \ge \sqrt{x}-1, \end{align*}so $a-c \ge \sqrt{x-y} \ge \sqrt{x}-1$, as desired. $\square$

WLOG $c \geq b \geq a$ Let $x = \lvert a-b \rvert$, $y = \lvert b-c \rvert$ and $x+y = \lvert c-a \rvert$. Subtracting one from both sides, multiplying by $2$ we get $$\frac{2(x^2+y^2+xy)}{ab+bc+ca} = \frac{x^2+y^2+(x+y)^2}{ab+bc+ca} = ab+bc+ca-1$$. But now note that, $ab+bc+ca = a^2+b^2+c^2-\frac{x^2+y^2+(x+y)^2}{2} = (a^2+b^2+c^2)-(x^2+y^2+xy)$. But now we assume the inequality is false. Then, $a^2+b^2+c^2 > (1+x+y)^2 = x^2+y^2+2xy+2x+2y+1$. So, $ab+bc+ca > xy+2x+2y+1$, so we have that $2(x^2+y^2+xy) > (xy+2x+2y)(xy+2x+2y+1) \geq (2x+2y)^2$ which is obviously false.

Define $x = a+b+c$ and $y = ab+bc+ca$. Our condition can be rewritten as \[2x^2 = y^2+5y \ge 6y \implies y \ge 1.\] Notice that our removing absolute values will always weakly decrease an expression. Thus \[\left(\frac{\sum |a-b|}{2}\right)^2 \ge \frac{\sum ((a-b)^2 + 2(a-b)(a-c))}{4} = x^2-3y.\] Hence our inequality is equivalent to \[x^2-2y \ge \left(\sqrt{x^2-2y}-1\right)^2 \iff 4x^2 \ge y^2+10y+1\]\[\iff 2(y^2+5y) \ge y^2+10y^+1 \iff y^2 \ge 1. \quad \blacksquare\]

Assume wlog that $a\leq b\leq c$ then we only need to prove $$b^2+2a(c+1)\leq 2c+1.$$ First consider the case $c\leq 5$. Define the functions $$f(a,b,c)=2c+1-b^2-2ac-2a, \qquad g(a,b,c)=2\left (a^2+b^2+c^2 \right)-(ab+bc+ca+1)(ab+bc+ca)$$for $a,b,c\in \{a,b,c\in [0,5]: g(a,b,c)=0\}$. In particular, $f$ achieves a global minimum. Note that $f>0$ at the endpoints, and so we will look at the interior. By Lagrange Multipliers we get \begin{align*} -2c-2 & =\lambda \left (4a-b-c-2bc^2-2b^2c-2ac^2-2ab^2-4abc \right) \\ -2b & =\lambda \left (4b-a-c-2ac^2-2a^2c-2bc^2-2ba^2-4abc\right) \\ 2-2a & =\lambda \left(4c-b-a-2ab^2-2a^2b-2cb^2-2ca^2-4abc\right). \end{align*}From this we find (by subtraction) $$\frac{2b-2c-2}{a-b}=\frac{-2b-2+2a}{b-c}=\frac{4-2a+2c}{c-a}=\lambda\left (2ab+2ac+2bc+5\right ).$$And this implies $b=\frac{a+c}2$. Let $x=\frac{c-a}2$ and $y=ac$. Then, $$0=2\left (a^2+\left (\frac{a+c}2\right)^2+c^2\right )-\left(a\frac{a+c}2+\frac{a+c}2 c+ca+1\right)\left(a\frac{a+c}2+\frac{a+c}2 c+ca\right)=-9y^2+y\left (3-12x^2\right )-4x^4+8x^2.$$Solving for $y$, we get $y=\frac 16 \left (1-4x^2\pm \sqrt{24x^2+1}\right)$. Now we can write everything in terms of $x$, $$2c+1-b^2-2ac-2a=4x+1-x^2-3y=x^2+4x+\frac 12\pm \frac 12 \sqrt{24x^2+1}$$and it's quite simple to verify the RHS is non negative, given $x\geq 0$. So we're done! Now suppose $c>5$. The given condition implies $$0=a^2\left (2-(b+c)^2\right )-a(2bc+1)(b+c)+2b^2+2c^2-b^2c^2-bc \qquad (\star)$$Note that $2-(b+c)^2<0$ and $-(2bc+1)(b+c)<0$, so for $(\star)$ to have a positive root we must have $$2b^2+2c^2-b^2c^2-bc\geq 0.$$From this it's easy to see that $b\leq \sqrt 2$. Now let's assume that $a>3/4$ and so $3/4<a,b\leq \sqrt 2$. Using this on $(\star)$ we get $$0<\left(\frac 34\right)^2\left(2-\left(c+\frac 34\right)^2\right)-\frac 34 \left (2\cdot \frac 34c+1\right)\left(c+\frac 34\right)+2c^2+2\left(\sqrt 2\right)^2-\left (\frac 34\right)^2c^2-\frac 34 c$$and this implies $\frac 14 c^2+\frac{51}{16}c-\frac{1087}{256}<0$. But we can easily check this inequality is false for $c>5$, and so we get a contradiction. Thus it must be that $a\leq \frac 34$. And so, $$b^2+2a(c+1)\leq 2+\frac 32 (c+1)\leq 2c+1,$$as desired.