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

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

Answer 1

@User88463: For example, look at the term $a^2\sqrt{b^2-bc+c^2}=a\sqrt{a^2(b^2-bc+c^2)}$. Then you simply apply AM-GM to the term $\sqrt{a^2(b^2-bc+c^2)}$: In particular, since you have $2$ non-negative terms, which are $a^2 $ and $b^2-bc+c^2$ respectively, by applying AM-GM to these $2$ terms, you will obtain: $$\frac{a^2+(b^2-bc+c^2)}{2} \ge \sqrt{a^2(b^2-bc+c^2)} \iff \sqrt{a^2(b^2-bc+c^2)} \le \frac{a^2+(b^2-bc+c^2)}{2} $$

Which is simply your AM-GM with $2$ variables. If you still cannot see it, let $A=a^2$ and $B=b^2-bc+c^2$, so in reality you actually have $\frac{A+B}{2} \ge \sqrt{AB}$. Multiplying by $a$ on both sides of your inequality, you will obtain: $$a\sqrt{a^2(b^2-bc+c^2)} \le \frac{a(a^2+b^2+c^2-bc)}{2} $$

Thus: $$\sum_{cyc} a^{2} \sqrt{b^{2}-b c+c^{2}}=\sum_{cyc}a \sqrt{a^{2}\left(b^{2}-b c+c^{2}\right)} \le \frac{1}{2} \sum_{cyc} a\left(a^{2}+b^{2}+c^{2}-b c\right)$$

Answer 2

Hint: Show that $b^2 - bc + c^2 \geq 0$.

There are lots of ways, like checking discriminant, completing the square, AM-GM.

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Oh come on. Not again. You really need to think these through.
Even if you're stuck, explain what you've tried and where you're stuck.

Answer 3

$$a\cdot (a\sqrt {b^2-bc+c^2}) \le \dfrac {1}{2}a (a^2+b^2+c^2-abc), ...$$

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

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

Which is Shur

Answer 4

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.