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$f:\mathbb{R}^k \rightarrow \mathbb{R}$ is differentiable function such that $\sum_{i=1}^{k} x_i \frac{df}{dx_i}(\mathrm{x} )\ge 0$ for $\mathrm{x}=(x_1,x_2,...,x_k) \in \mathbb{R} ^k$. Prove that function f is bounded below.

All what I've got: $\sum_{i=1}^{k} x_i \frac{df}{dx_i}(\mathrm{x} )= (x_1,x_2,...,x_k) \cdot (\nabla f(x))=\nabla _w f(\mathrm{x})$, hence $\nabla _w f(\mathrm{x}) \ge 0$. I think that it's important remark, but I don't know how to end this solution.

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A sketch of solution. You'll have to fill in the gaps.

Try fixing $x\ne 0$ and evaluating $f$ along the half-line $\{tx\ :\ t\ge 0\}$. As you have already noticed, the resulting function of $t$ is nondecreasing. This means that, whatever $y\in \mathbb{R}^k$ you take outside of a fixed closed ball (say, of radius one) centered at the origin, you will have $$f(y)\ge f(\text{some point inside the ball}).$$ Taking infimums, you can conclude that $$\inf_{\mathbb{R}^k} f(y)=\inf_{\{\lvert y \rvert \le 1\}} f(y).$$ Now you only have to use the continuity of $f$.

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Let $x \in \mathbb R^k$, consider the function $g \colon [0,1] \to \mathbb R$, given by $g(t) = f(tx)$. Then $g$ is differentiable and we have $g'(t) = f'(tx)x$, so for $t > 0$ we have $g'(t) = \frac 1t f'(tx)tx = \frac 1t\sum_{i=1}^k\frac{\partial f}{\partial x_i}(tx)tx_i \ge 0$. Hence \begin{align*} f(x) &= g(1)\\ &= g(0) + \int_0^1 g'(t)\,dt\\ &\ge g(0)\\ &= f(0) \end{align*} So $f$ is bounded below by $f(0)$.

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