# Prove that $b(c+a)<\sqrt{2}$ for $a^2+b^2+c^2-a(b+c)=2-\sqrt{2}$

Let $$a,b,c$$ be real numbers with $$a^2+b^2+c^2-a(b+c)=2-\sqrt{2}$$. Prove that $$b(c+a)<\sqrt{2}$$.

I've tried a lot of ways to use AM-GM on the inequality and to try to prove equivalent inequalities but I can't be able to solve it. Any ideas?

• A peanut (ellipsoid) between two crushing walls (hyperbolic cylinder). How comfortable are you with quadric surfaces? Commented May 13, 2023 at 19:22
• @CatalinZara Was looking for a more elementary proof, but if it works then sure! Commented May 13, 2023 at 19:27
• Change coordinates to get one of the conditions in a more standard form. Try a linear change of variables (such as $a=\alpha x + \beta y + \gamma z$, and so on for $b$ and $c$) that would eliminate the $ab$, $ac$, and $bc$ terms in one of the conditions. Commented May 13, 2023 at 19:34
• @CatalinZara Do you have any tips on finding such transformations? Commented May 13, 2023 at 19:50
• Use the following fact. $ax^2+bx+c\geq0$, where $a\neq0$, for any real $x$ iff $a>0$ and $b^2-4ac\leq0.$ Commented May 13, 2023 at 20:18

Remarks: There are many proofs. Here is one.

If $$b = 0$$, clearly the desired inequality is true.

In the following, assume that $$b \ne 0$$.

The desired inequality is written as $$b(c + a)(\sqrt 2 - 1) < (\sqrt 2 - 1)\sqrt 2 = 2 - \sqrt 2$$ or $$b(c + a)(\sqrt 2 - 1) < a^2 + b^2 + c^2 - a(b + c)$$ or (simply completing the squares in $$a$$ and then $$c$$) $$\frac14(2a - b\sqrt 2 - c)^2 + \frac34 \left(c - \frac{3\sqrt 2 - 2}{3}b\right)^2 + \frac{3\sqrt 2 - 4}{3} b^2 > 0$$ which is true.

We are done.

• Could you explain more about how you completed the squares and flipped the sign of the inequality? Commented May 14, 2023 at 2:32
• @TiagoCavalcante Since $A < B$ is equivalent to $B > A$. So $b(c + a)(\sqrt 2 - 1) < a^2 + b^2 + c^2 - a(b + c)$ is equivalent to $a^2 + b^2 + c^2 - a(b + c) > b(c + a)(\sqrt 2 - 1)$. The completing procedure is quite common: for example $a^2 + pa + q = (a + p/2)^2 + q - p^2/4$. Commented May 14, 2023 at 2:36

Say we have $$B$$ a positive definite matrix and $$A$$ and symmetric matrix ( of same size). Now we want the maximal and the minimal values of

$$v \mapsto \frac{\langle A v, v \rangle } {\langle B v, v\rangle }$$

In many cases one treats the case $$B = I$$, the identity matrix. However, we can reduce to this case in this way. Write

$$\langle B v, v\rangle = \langle B^{\frac{1}{2}} v, B^{\frac{1}{2}} v\rangle = \langle w, w \rangle \\ \langle A v, v\rangle = \langle A B^{-\frac{1}{2}} w, B^{-\frac{1}{2}} w\rangle =\langle B^{-\frac{1}{2}} A B^{-\frac{1}{2}} w, w\rangle$$ so $$\max \frac{\langle A v, v \rangle } {\langle B v, v\rangle } = \max \frac{\langle B^{-\frac{1}{2}} A B^{-\frac{1}{2}} w, w\rangle}{\langle w, w \rangle}$$

and that equals the largest of the eigenvalues of $$B^{-\frac{1}{2}} A B^{-\frac{1}{2}}$$, that is, the largest of the root of the polynomial

$$\det(t I- B^{-\frac{1}{2}} A B^{-\frac{1}{2}}) = \det B^{-1}\det ( t B - A)$$

Now in our case we have

$$\begin{eqnarray} b(a+c)&=& (a,b,c) \cdot A \cdot (a,b,c)^t\\ a^2 + b^2 + c^2 -a(b+c) &=& (a,b,c) \cdot B \cdot (a,b,c)^t \end{eqnarray}$$

where $$\begin{eqnarray} A &=& \left ( \begin{matrix} 0 & \frac{1}{2} & 0 \\ \frac{1}{2} & 0 & \frac{1}{2} \\ 0 & \frac{1}{2} & 0 \end{matrix} \right) \\ B &=& \left ( \begin{matrix}1&-\frac{1}{2} & -\frac{1}{2}\\ -\frac{1}{2}& 1& 0 \\ -\frac{1}{2} & 0 & 1 \end{matrix} \right) \end{eqnarray}$$

Now we have $$\det ( t B - A) = \frac{1}{2} t^3 - \frac{3}{4} t^2 - \frac{3}{4} t$$ with roots $$0, \frac{1}{4}( 3 \pm \sqrt{33})$$. Therefore

$$\max_{(a,b,c)\ne (0,0,0)} \frac{b(c+a)}{ a^2 + b^2 + c^2 - a(b+c)}= \frac{1}{4}(3 + \sqrt{33})$$

Note that $$\frac{1}{4}(3 + \sqrt{33}) < \frac{\sqrt{2}}{2 - \sqrt{2}}$$.

$$\bf{Added:}$$ We can get the upper bound from the OP as follows: maximize $$\frac{b(a+c)}{a^2 + b^2 + c^2}$$, then $$\frac{a^2+b^2 + c^2}{ a^2 + b^2 + c^2 - a(b+c)}$$ and take the product to get the result. This in fact quite simple.

$$a^2 +b^2 + c^2 \ge b^2 + \frac{(a+c)^2}{2} \ge 2 b \cdot \frac{a+c}{\sqrt{2}}$$ so $$\begin{eqnarray} \frac{b(a+c)}{a^2+b^2+c^2} &\le &\frac{1}{\sqrt{2}}\\ \frac{a^2 + b^2 + c^2}{a^2 + b^2 + c^2 - a(b+c)} &\le &\frac{1}{1-\frac{1}{\sqrt{2}}} \end{eqnarray}$$ so by multiplication $$\frac{b(a+c)}{a^2 + b^2 + c^2 - a(b+c)} \le \frac{1}{\sqrt{2}-1} = \frac{\sqrt{2}}{2- \sqrt{2}}$$

Now the inequality is strict because the above inequalities are not satisfied simultaneously.

$$\bf{Added:}$$ Let us prove "by hand" that we have the inequality

$$\frac{1}{4}(3 + \sqrt{33})\ge \frac{b(a+c)}{ a^2 + b^2 + c^2 -a(b+c)}$$ indeed LHS $$-$$ RHS is a positive multiple of $$(3 + \sqrt{33})a^2 - [ (7 + \sqrt{33}) b + (3 +\sqrt{33})c]\cdot a + [ (3+\sqrt{33})(b^2+c^2) - 4 bc]$$ which has discriminant in $$a$$ a negative multiple of $$(16 b -3(1 + \sqrt{33})c)^2$$ so the expression is $$\ge 0$$ with possible equality.