Proof of one inequality $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$

Solution 1:

Using Cauchy-Schwarz Inequality twice:

$a^4 + b^4 +c^4 \geq a^2b^2 +b^2c^2 +c^2a^2 \geq ab^2c +ba^2c +ac^2b = abc(a+b+c)$

Solution 2:

I have come up with an answer with myself. Using CS inequality $$(a^4+b^4+c^4)(1+1+1)\geq(a^2+b^2+c^2)^2$$ $$(a^2+b^2+c^2)(1+1+1)\geq(a+b+c)^2$$ Hence we have $$a^4+b^4+c^4\geq\frac{(a+b+c)^4}{27}=(a+b+c)\left(\frac{a+b+c}{3}\right)^3\geq abc(a+b+c)$$

Solution 3:

Here other two answers used Cauchy-Scwartz Inequality. I am giving a simple $AM\ge GM$ inequality proof.

You asked, $$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\ge a+b+c\\\implies a^4+b^4+c^4\ge a^2bc+b^2ca+c^2ab$$

Now, from, $AM\ge GM$, we have $$\frac {a^4+ a^4+b^4+c^4}4\ge \left(a^4\cdot a^4\cdot b^4\cdot c^4\right)^{1/4}=a^2bc\tag 1$$

Similarly, $$\frac {a^4+ b^4+b^4+c^4}4\ge \left(a^4\cdot b^4\cdot b^4\cdot c^4\right)^{1/4}=ab^2c\tag 2$$ and also, $$\frac {a^4+ b^4+c^4+c^4}4\ge \left(a^4\cdot b^4\cdot c^4\cdot c^4\right)^{1/4}=abc^2\tag 3$$

Now, summing up $(1),(2),(3)$, we have, $a^4+b^4+c^4\ge a^2bc+b^2ca+c^2ab$, that is $$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\ge a+b+c$$

Solution 4:

By Holder $$\sum_{cyc}\frac{a^3}{bc}\geq\frac{(a+b+c)^3}{3(ab+ac+bc)}=\frac{(a+b+c)\cdot(a+b+c)^2}{3(ab+ac+bc)}\geq a+b+c$$

Solution 5:

Another way.

By Rearrangement $$\sum_{cyc}\left(\frac{a^3}{bc}-a\right)=\sum_{cyc}\frac{a}{bc}(a^2-bc)\geq\frac{1}{3}\sum_{cyc}\frac{a}{bc}\sum_{cyc}(a^2-bc)=\frac{1}{6}\sum_{cyc}\frac{a}{bc}\sum_{cyc}(a-b)^2\geq0.$$