Having $x_2>\frac{1}{x_1}$, $y_2>\frac{1}{y_1}$, and $x_1, y_1>0$ can we bound $[(1-c)x_2+cy_2]^2+[(1-c)x_1+cy_1]^2$ below from $2$ ?
(We get $c$ from the formula of the mean value for functions in $\mathbb{R}^2$.)
I thought to do something like:
\begin{align*}[(1-c)x_2+cy_2]^2+[(1-c)x_1+cy_1]^2&>\left [\frac{(1-c)}{x_1}+\frac{c}{y_1}\right ]^2+[(1-c)x_1+cy_1]^2\\ & >\left (\frac{1-c}{x_1}\right )^2+\left (\frac{c}{y_1}\right )^2+\left ((1-c)x_1\right )^2+\left (cy_1\right )^2 \\ & > \frac{(1-c)^2}{x_1^2}+\frac{c^2}{y_1^2}+(1-c)^2x_1^2+c^2y_1^2 \\ & >\frac{(1-c)^2}{x_1^2}+\frac{c^2}{y_1^2}\end{align*}
But I don't know if this helps.
Could you give me a hint?
I'd start like this $$[(1-c)x_1+cy_1]^2+[(1-c)x_2+cy_2]^2\geq2[\frac{(1-c)x_1+cy_1+(1-c)x_2+cy_2}{2}]^2$$ You can come up with this inequality drawing a parabola, and connecting points on the parabola. It's known as Jensen's inequality. It's simple, and beautiful at the same time.
You can then do something like this, assuming that $c\in[0, 1]$. $$ 2[\frac{(1-c)x_1+cy_1+(1-c)x_2+cy_2}{2}]^2>2[\frac{(1-c)(x_1+\frac{1}{x_1})+c(y_1+\frac{1}{y_1})}{2}]^2\geq 2[\frac{2(1-c)+2c}{2}]^2=2 $$ This comes from $x_2>\frac{1}{x_1}$ and $y_2>\frac{1}{y_1}$, and then from a fact that for $x>0$ we have: $$\frac{1}{x}+x\geq 2$$ I realized i read wrong, and thought there was $x_2=\frac{1}{x_1}$, $y_2=\frac{1}{y_1}$, if i saw it was an inequality, i would of added more details.