Assume A is a symmetric and positive definite matrix.Show that the function $f:\mathbb{R}^{n}\rightarrow\mathbb{R}$, $f(u):=\frac{1}{2}u^{\top}Au-b^{\top}u+2\sum_{i=1}^{n}\cosh(u_{i})$ satisfies the following relation: $\forall u,v\in\mathbb{R}^{n},u\neq v,λ\in(0,1),$
$$λf(u)+(1-λ)f(v)-f(λu+(1-λ)v)>\frac{1}{2}λ(1-λ)\left\Vert u-v\right\Vert _{A}^{2}$$
Attempt: I know that $\left\Vert u-v\right\Vert _{A}^{2}=(u-v)^{\top}A(u-v)$ and tried to prove that
$$λf(u)+(1-λ)f(v)-f(λu+(1-λ)v)>\frac{1}{2}λ(1-λ)\left\Vert u-v\right\Vert _{A}^{2}$$$$= λf(u)+(1-λ)f(v)-f(v+λ(u-v))-\frac{1}{2}λ(1-λ)\left\Vert u-v\right\Vert _{A}^{2}>0$$ but the calculations are very complex and I don't know if I have to use a "trick" to solve it. I have also proved that $g(u)=\frac{1}{2}u^{\top}Au-b^{\top}u$ is a quadratic form and it is a strictly convex-with means that for $λ\in(0,1)$
$$λg(u)+(1-λ)g(v)>g(v+λ(u-v))$$