# Lower bound for Strongly convex and Lipshcitz gradient function

I was reading this paper but could not understand one of it's condition. It says a function $$f(x)$$ is twice differentiable and strongly convex with parameter $$m$$ and Lipschitz continuous gradient with parameter $$M$$ and $$M\geq m$$. Then: $$(x-y)^T(\nabla f(x)-\nabla f(y) \geq \frac{mM}{m+M} \lVert x-y \rVert^2+\frac{1}{m+M}\lVert \nabla f(x)-\nabla f(y) \rVert ^2$$

I know that if $$f$$ is strongly convex and Lipshcitz continuous gradient, then : $$(x-y)^T(\nabla f(x)-\nabla f(y)) \geq m \rVert x-y\lVert ^2$$ and $$(x-y)^T(\nabla f(x)-\nabla f(y)) \geq \frac{1}{M} \lVert \nabla f(x)-\nabla f(y) \rVert ^2$$. I tried multiply each by $$M$$ and $$m$$ and adding up but that didn't quite work.

Are you sure about the coefficient within the second term on the right hand side?

As you mentioned the strong convexity means

$$\left( x-y | \nabla f(x) - \nabla f(y) \right) \ge m ||x-y||^2$$

(I write $$\left(.|.\right)$$ for the inner product). And the Lipschitz property of the gradient means

$$||\nabla f(x) - \nabla f(y)|| \le M ||x-y||.$$

Combining both gives:

$$\left( x-y | \nabla f(x) - \nabla f(y) \right) = m\frac{m+M}{m+M} ||x-y||^2 \ge \frac{mM}{m+M} ||x-y||^2 + \frac{m^2}{m+M} ||x-y||^2 \ge \frac{mM}{m+M} ||x-y||^2 + \frac{m^2}{M^2(m+M)} ||\nabla f(x) - \nabla f(y)||^2.$$