Prove that $a^{2} \sqrt{b^{2}-b c+c^{2}}+b^{2} \sqrt{c^{2}-c a+a^{2}}+c^{2} \sqrt{a^{2}-a b+b^{2}} \leq a^{3}+b^{3}+c^{3}$ Question -


  Let $a, b, c$ be non-negative real numbers. Prove that
  $$
a^{2} \sqrt{b^{2}-b c+c^{2}}+b^{2} \sqrt{c^{2}-c a+a^{2}}+c^{2} \sqrt{a^{2}-a b+b^{2}} \leq a^{3}+b^{3}+c^{3}
$$


my doubt -

author writes

Applying AM-GM inequality, we have

$\sum_{c y c} a^{2} \sqrt{b^{2}-b c+c^{2}}=a \sqrt{a^{2}\left(b^{2}-b c+c^{2}\right)} \leq \frac{1}{2} \sum_{c y c} a\left(a^{2}+b^{2}+c^{2}-b c\right)$

but how we can apply am-gm if there is negative term inside ??? can someone clear this step that how they apply am-gm and got this result..

thankyou
As pointed out by others, why do you even assume $b^2-bc+c^2$ can be negative?

Simply write it as $(b-c)^2+bc$. The first term, being a perfect square, is clearly always non-negative, while the second term, being a product of two non-negative numbers ( as per the question), must also be non-negative. Thus their sum is also non-negative.
$$a\cdot (a\sqrt {b^2-bc+c^2}) \le \dfrac {1}{2}a (a^2+b^2+c^2-abc), ...$$

$$LHS \le \dfrac {1}{2}(a^3+b^3+c^3)+\dfrac {1}{2}\sum_{cyc}a^2(b+c)-\dfrac {3}{2}abc \le a^3+b^3+c^3$$

$$a^3+b^3+c^3+3abc\ge \sum_{cyc}a (b^2+c^2)$$

Which is Shur