# Application of Cauchy Schwarz Inequality

I'm arriving at the final step of a pointwise estimate of the heat kernel on a Riemannian manifold but there's one step where I think I don't see the trick. It is said that the following identity follows from the Scwarz inequality $$\left(r - \sqrt{t}\right)^2 \geq \frac{r^2}{1+c}- \frac{t}{c}$$

I've tried playing with some weights but I can't seem to get this to fall out. All the variables are positive numbers and $r^2 > 4t$

• can you give us some Information about the variables? – Dr. Sonnhard Graubner Feb 16 '17 at 19:59
• I originally had a typo that I fixed. All quantities are positive and $r^2 > 4t$ – Clown Baby Feb 16 '17 at 20:00

making some steps backwards then we get after squaring $$r^2-\frac{r^2}{1+c}+t+\frac{t}{c}\geq 2r\sqrt{t}$$ and this is equaivalent to $$r^2\frac{c}{1+c}+t\frac{1+c}{c}\geq 2r\sqrt{t}$$ squaring again we obtain $$r^4\left(\frac{c}{1+c}\right)^2+t^2\left(\frac{1+c}{c}\right)^2+2r^2t\geq 4r^2t$$ putting Things together we get $$\left(r^2\frac{c}{1+c}-t\frac{1+c}{c}\right)^2\geq 0$$ without Cauchy Bunjakowski Schwartz