Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2+abc=4$. How do you prove that $\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge a+b+c+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$?
My Approach: I tried basic algebraic techniques, but they didn't work. So, I moved on to a trigonometric approach. Note that $a^2+b^2+c^2+abc=4$ implies that $a<2$, and similarly for $b$ and $c$. So, I can let
$\qquad a=2\cos\alpha \qquad b=2\cos\beta \qquad c=2\cos\gamma \qquad 0<\alpha,\beta,\gamma<90^{\circ}$
where $a,b,c$ are side lengths of an acute triangle. These are the lengths of acute triangle because if any one of the angle was to be greater than or equal to $90^{\circ}$, one of the side would be zero or negative due to our assumption. (By the way, if we put supposed values of $a,b,c$ in L.H.S of the given equation, the result is equal to $4$.)
Now, we need to prove that
$\dfrac{\cos\alpha+\cos\beta}{\cos\gamma}+\dfrac{\cos\beta+\cos\gamma}{\cos\alpha}+\dfrac{\cos\gamma+\cos\alpha}{\cos\beta}\ge 2(\cos\alpha+\cos\beta+\cos\gamma)+\dfrac{1}{2}\left(\dfrac{1}{\cos\alpha}+\dfrac{1}{\cos\beta}+\dfrac{1}{\cos\gamma}\right)$
which can be further simplified into
$\left(\cos\alpha+\cos\beta+\cos\gamma-\dfrac{1}{2}\right)\left(\dfrac{1}{\cos\alpha}+\dfrac{1}{\cos\beta}+\dfrac{1}{\cos\gamma}-2\right)\ge4$.
We need to prove this inequality given that $\alpha+\beta+\gamma=180^{\circ}$ and $0<\alpha,\beta,\gamma<90^{\circ}$.
I tried to look for triangle inequalities which would help me out but was too overwhelmed and frustrated after going through Wikipedia.
After struggling a lot with this question, I'm not even sure if trigonometric approach is the best option. Please help me out!
Best Answer
By AM-GM $$4=a^2+b^2+c^2+abc=a^2+b^2+c^2+\sqrt{a^2b^2c^2}\leq a^2+b^2+c^2+\sqrt{\left(\frac{a^2+b^2+c^2}{3}\right)^3}.$$
Now, let $a^2+b^2+c^2=3x^2,$ where $x>0$.
Thus, $$4\leq3x^2+x^3$$ or $$x^3+3x^2-4\geq0$$ or $$x^3-x^2+4x^2-4x+4x-4\geq0$$ or $$(x-1)(x^2+4x+4)\geq0$$ or $$x\geq1$$ or $$\frac{a^2+b^2+c^2}{3}\geq1,$$ which gives $$a^2+b^2+c^2\geq3.$$ In another hand, it's obvious (for exampleб by the Carnot's theorem) that $$a+b+c\leq3.$$ Id est, $$1\leq\sqrt{\frac{a^2+b^2+c^2}{3}},$$ $$a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq3+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\sqrt{\frac{a^2+b^2+c^2}{3}},$$ which says that it's enough to prove that: $$\sum_{cyc}\frac{a+b}{c}\geq3+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\sqrt{\frac{a^2+b^2+c^2}{3}},$$ which is homogeneous already.
Can you end it now?
I got that the last inequality is true even for any positives $a$, $b$ and $c$.
Indeed, we need to prove that: $$\sum_{cyc}(a^2b+a^2c-abc)\geq(ab+ac+bc)\sqrt{\frac{a^2+b^2+c^2}{3}}.$$ Now, let $a+b+c=3u$ and $ab+ac+bc=3v^2$, where $v>0$.
Thus, by AM-GM $abc\leq v^3$ and since $$\sum_{cyc}(a^2b+a^2c)=(a+b+c)(ab+ac+bc)-3abc=9uv^2-3abc,$$ it's enough to prove that: $$9uv^2-6v^3\geq3v^2\sqrt{3u^2-2v^2}$$ or $$3u-2v\geq\sqrt{3u^2-2v^2}$$ or $$\sqrt{3u^2-2v^2+6(u-v)^2}\geq\sqrt{3u^2-2v^2}$$ and we are done!