Suppose $\varphi \in C^{\infty}(\mathbf{R}^n)$. I would like to show that $$ \sum_{j=1}^n \| \partial_j \varphi \|_{L^{\infty}} \leq 2n\big(\|\varphi \|_{L^{\infty}}\big)^{\frac{1}{2}} \bigg(\sum_{j=1}^n \sum_{k=1}^n \|\partial_j \partial_k \varphi \|_{L^{\infty}}\bigg)^{\frac{1}{2}}. $$ I'm positive that this has to do with the expansion $$ \varphi(x+h) - \varphi(x) = \nabla \varphi(x) \cdot h + \frac{1}{2}\sum_{j,k=1}^n \partial_j \partial_k \varphi(\xi) h_j h_k $$ given by the multivariable version of Taylor's theorem. Also, in order to obtain to factor of $2n$ and multiplication of square roots, I'll probably want to show each of $$ \| \partial_j \varphi \|_{L^{\infty}} \leq 2 \|\varphi \|_{L^{\infty}} $$ and $$ \| \partial_j \varphi \|_{L^{\infty}} \leq 2 \sum_{i=1}^n \sum_{k=1}^n \|\partial_i \partial_k \varphi \|_{L^{\infty}} $$ separately for all $j$, because then we get $$ \bigg(\sum_{j=1}^n \| \partial_j \varphi\|_{L^\infty}\bigg)^{1/2} \leq \left(2n \|\varphi\|_{L^\infty}\right)^{1/2} $$ and $$ \bigg(\sum_{j=1}^n \| \partial_j \varphi\|_{L^\infty}\bigg)^{1/2} \leq \left(2n \sum_{i=1}^n \sum_{k=1}^n \|\partial_i \partial_k \varphi \|_{L^{\infty}}\right)^{1/2} $$ which by multiplying each side together gives the desired result. I'm not entirely certain if this is the right approach though. Any hints would be helpful. I don't absolutely need a full solution, but I won't mind if you give one. Thanks!
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