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

Question

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?

Answer 1

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.

Answer 2

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.