Prove that $\left(a^2b+b^2c+c^2a\right)\left(\dfrac{1}{\sqrt{1+a^2}}+\dfrac{1}{\sqrt{1+b^2}}+\dfrac{1}{\sqrt{1+c^2}}\right)\leq \dfrac{3}{2}.$

Question

Let $a,b,c$ be nonnegative real numbers such that $a^2+b^2+c^2=1$. Prove that$$\left(a^2b+b^2c+c^2a\right)\left(\dfrac{1}{\sqrt{1+a^2}}+\dfrac{1}{\sqrt{1+b^2}}+\dfrac{1}{\sqrt{1+c^2}}\right)\leq \dfrac{3}{2}.$$

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I have no idea for this. How to handle with kind of assumption like $a^2+b^2+c^2=1$? Please help me, thank you very much.