Suppose $A$ is positive-definite, symmetric and real. Show that for $u,v\in\mathbb{R}^{n}$ that $$2|Au\cdot v|\leq\rho^{\frac{1}{2}}(u\cdot Au+|v|^{2}),$$ where $\rho$ is the maximal eigenvalue of $A$, e.g. $\rho:=\text{max}(\rho_{1},\ldots,\rho_{n}).$ I've tried fiddling around with different things, but haven't arrived at a concise proof, and I need the result for some other things that I am doing right now, so I figured I'd see if anyone on here had a proof in mind.
Edit (1) (Response to Gary)
At first try, this leads to something similar to what I already obtained by using the identity $||A||_{2}=\sqrt{\rho(AA^{t})}=\sqrt{\rho(A^{2})}=\rho$ (since $A=A^{T}$). In details, $$2|Au\cdot v|\leq2|Au||v|\leq2||A||_{2}|u||v|\leq2\rho|u||v|\leq\rho^{\frac{1}{2}}(|u|^{2}+|v|^{2}).$$ Similarly, following Gary's suggestion, if $A$ is diagonalized and we're in the orthonormal eigenbasis, we have $$2|Au\cdot v|=2\left|\sum\limits_{j=1}^{n}\rho_{j}u_{j}v_{j}\right|\leq2\rho(u\cdot v)\leq2\rho|u||v|\leq\rho^{\frac{1}{2}}(|u|^{2}+|v|^{2}),$$ which is the same estimate as above.
In any case, the proof is reduced to showing $$|u|^{2}\leq u^{T}Au=\sum\limits_{j=1}^{n}u_{j}\rho_{j}u_{j}=\sum\limits_{j=1}^{n}\rho_{j}u_{j}^{2}\leq\rho|u|^{2}.$$
But this inequality can hold if and only if $\rho\geq1$. Is there a theorem which concludes positive definite, real and symmetric matrices necessarily have an eigenvalue greater than or equal to $1$? I doubt this, so either I've made a mistake or the problem that I am working on left out an assumption on $A$...thoughts? Is there a sharper estimate than just the Cauchy inequality used above?
