I found this problem and its solution in a FB group dedicated to maths. Here they are.

**Problem**:

If $a,b,c \gt 0$ prove that

$$\frac{a^2 + b^2}{c} + \frac{b^2 + c^2}{a} + \frac{c^2 + a^2}{b} \ge 2(a+b+c)$$

**Solution:**

By the AM-GM inequality we have that

$$\frac{a^2}{c} + c + \frac{b^2}{c} + c \ge 2\sqrt{a^2} + 2\sqrt{b^2} = 2(a+b)$$

$$\frac{b^2}{a} + a + \frac{c^2}{a} + a \ge 2\sqrt{b^2} + 2\sqrt{c^2} = 2(b+c)$$

$$\frac{c^2}{b} + b + \frac{a^2}{b} + b \ge 2\sqrt{c^2} + 2\sqrt{a^2} = 2(c+a)$$

When we sum up these 3 inequalities we get the desired inequality.

**Note:** This solution is due to Ghimisi Dumitrel, Professor at Colegiul National Ionita Asan, Caracal.

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