Inequality for Olympiad students

Solution 1:

Another way (L.Hadassy, Y.Ilany).

If $a\geq b\geq c$ we have $$\sqrt{3a+\frac{1}{b}}+\sqrt{3b+\frac{1}{c}}\geq\sqrt{3a+\frac{1}{c}}+\sqrt{3b+\frac{1}{b}}$$ because it's $$(a-b)(b-c)\geq0.$$ Thus, $$\sum_{cyc}\sqrt{3a+\frac{1}{b}}\geq\sqrt{3a+\frac{1}{c}}+\sqrt{3c+\frac{1}{a}}+\sqrt{3b+\frac{1}{b}}.$$ If $a\geq c\geq b$ we have $$\sqrt{3b+\frac{1}{c}}+\sqrt{3c+\frac{1}{a}}\geq\sqrt{3b+\frac{1}{a}}+\sqrt{3c+\frac{1}{c}}$$ because it's $$(a-c)(c-b)\geq0,$$ which gives $$\sum_{cyc}\sqrt{3a+\frac{1}{b}}\geq\sqrt{3a+\frac{1}{b}}+\sqrt{3b+\frac{1}{a}}+\sqrt{3c+\frac{1}{c}}.$$ Now we see that in any case we need to prove that: $$\sqrt{3a+\frac{1}{c}}+\sqrt{3c+\frac{1}{a}}+\sqrt{3b+\frac{1}{b}}\geq6$$ or $$\sqrt{3ac+1}\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{c}}\right)+\sqrt{3b+\frac{1}{b}}\geq6,$$ where $a\geq b\geq c$ or $c\geq b\geq a$.

Now, let $a+c=p=constant,$ $f(a,c)=\sqrt{3ac+1}\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{c}}\right)$ and $F(a,c,\lambda)=f(a,c)+\lambda(a+c-p).$

Thus, in the minimum point we have $$\frac{\partial F}{\partial a}=\frac{\partial F}{\partial c}=0,$$ which gives $$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial c}$$ or $$(\sqrt{a}-\sqrt{c})(3\sqrt{a^3c^3}-\sqrt{ac}-a-c)=0.$$ 1) $a=c$.

Thus, $b=3-2a,$ where $0<a<\frac{3}{2}$ and we need to prove that $$2\sqrt{\frac{3a^2+1}{a}}+\sqrt{9-6a+\frac{1}{3-2a}}\geq6,$$ which is true.

2) $3\sqrt{a^3c^3}-\sqrt{ac}-a-c=0.$

Let $\sqrt{ac}=x$.

Thus, $a+c=3x^3-x,$ $b=3+x-3x^3$ and since $0<b\leq\frac{3}{2},$ we obtain: $0.932...<x\leq1.11...$ and we need to prove that: $$\sqrt{3x^2+1}\cdot\sqrt{\frac{a+c+2\sqrt{ac}}{ac}}+\sqrt{3(3+x-3x^3)+\frac{1}{3+x-3x^3}}\geq6$$ or $$\frac{3x^2+1}{\sqrt{x}}+\sqrt{3(3+x-3x^3)+\frac{1}{3+x-3x^3}}\geq6,$$ which is true and it ends the proof.

Solution 2:

My solution begins by the same steps.

By C-S we obtain: $$\sum_{cyc}\sqrt{3a+\frac{1}{b}}=\sqrt{\sum_{cyc}\left(3a+\frac{1}{b}+2\sqrt{\left(3a+\frac{1}{b}\right)\left(3b+\frac{1}{c}\right)}\right)}=$$ $$=\sqrt{\sum_{cyc}\left(3+\frac{1}{a}+2\sqrt{\left(a+2a+\frac{1}{b}\right)\left(\frac{1}{c}+2b+b\right)}\right)}\geq$$ $$\geq \sqrt{\sum_{cyc}\left(3+\frac{1}{a}+2\left(\sqrt{\frac{a}{c}}+2\sqrt{ab}+1\right)\right)}= \sqrt{\sum_{cyc}\left(5+\frac{1}{a}+2\sqrt{\frac{a}{c}}+4\sqrt{ab}\right)}.$$ Thus, it's enough to prove that: $$\sum_{cyc}\left(\frac{1}{a}+2\sqrt{\frac{a}{c}}+4\sqrt{ab}\right)\geq21.$$

Now, we'll prove that for any positives $a$, $b$ and $c$ the following inequality holds. $$\frac{a}{c}+\frac{b}{a}+\frac{c}{b}+\frac{(4\sqrt2-3)(ab+ac+bc)}{a^2+b^2+c^2}\geq4\sqrt2.$$

Indeed, we need to prove that: $$\sum_{cyc}a^2b\sum_{cyc}a^2+(4\sqrt2-3)\sum_{cyc}a^2b^2c\geq4\sqrt2\sum_{cyc}a^3bc$$ or $$\sum_{cyc}(a^4b+a^3c^2-4\sqrt2a^3bc+(4\sqrt2-2)a^2b^2c)\geq0$$ or $$\sum_{cyc}b(a-b)^2(a-\sqrt2c)^2\geq0,$$ which ends a proof of the lemma.

Thus, since $\sum\limits_{cyc}(a^2-ab)\geq0,$ we obtain: $$\frac{a}{c}+\frac{b}{a}+\frac{c}{b}+\frac{2.5(ab+ac+bc)}{a^2+b^2+c^2}-5.5\geq$$ $$\geq\frac{a}{c}+\frac{b}{a}+\frac{c}{b}+\frac{(4\sqrt2-3)(ab+ac+bc)}{a^2+b^2+c^2}-4\sqrt2\geq0.$$ Thus, it's enough to prove that:

$$\sum_{cyc}\left(\frac{1}{a}+2\left(\frac{5.5}{3}-\frac{2.5\sqrt{ab}}{3}\right)+4\sqrt{ab}\right)\geq21$$ or $$\sum_{cyc}\left(\frac{3}{a}+7\sqrt{ab}-10\right)\geq0$$ or $$\sum_{cyc}\left(\frac{3}{a^2}+7ab-10\right)\geq0,$$ where $a$, $b$ and $c$ are positives such that $a^2+b^2+c^2=3.$

Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, $3u^2-2v^2=1$ and we need to prove that: $$\frac{9v^4-6uw^3}{w^6}\geq10(3u^2-2v^2)-7v^2$$ and since the condition does not depend on $w^3$, we need to prove that $f(w^3)\geq0,$

where $f$ is a concave function (the coefficient before $w^6$ is negative).

But the concave function gets a minimal value for an extreme value $w^3$,

which happens in the following cases.

1. $w^3\rightarrow0^+$. In this case our inequality is obvious;

2. Two variables are equal.

Let $b=c$.

Thus, after homogenization we can assume $b=c=1$ and it's enough to prove that: $$\frac{(a^2b^2+a^2c^2+b^2c^2)(a^2+b^2+c^2)}{a^2b^2c^2}+\frac{21(ab+ac+bc)}{a^2+b^2+c^2}\geq30$$ or $$\frac{(2a^2+1)(a^2+2)}{a^2}+\frac{21(2a+1)}{a^2+2}\geq30$$ or $$(a-1)^2(2a^4+4a^3-15a^2+8a+4)\geq0,$$ which is true because by AM-GM $$2a^4+4a^3-15a^2+8a+4\geq4\sqrt[4]{2a^4\cdot4a^3\cdot 8a\cdot4}-15a^2=a^2>0$$ and we are done!