Prove $\sqrt{\frac{a+b}{c+ab}}+\sqrt{\frac{c+b}{a+cb}}+\sqrt{\frac{a+c}{b+ac}}\ge 3,$ if $ab+bc+ca=3.$

Question

If $a,b,c>0: ab+bc+ca=3,$ then prove that$$\sqrt{\frac{a+b}{c+ab}}+\sqrt{\frac{c+b}{a+cb}}+\sqrt{\frac{a+c}{b+ac}}\ge 3.$$ I've tried to use AM-GM without success.

Indeed, we need to prove $$(a+b)(b+c)(c+a)\ge (ab+c)(bc+a)(ca+b),$$or $$3(a+b+c)-abc\ge abc+(abc)^2+abc(a^2+b^2+c^2)+(ab)^2+(bc)^2+(ca)^2,$$ or $$3(a+b+c)+4abc\ge (abc)^2+abc(a+b+c)^2+9-2abc(a+b+c).$$ I'm stuck to prove the last inequality is true.

Hope you help me continue my idea. Thanks for your interest.

Answer 1

The inequality $$(a+b)(a+c)(b+c)\geq(ab+c)(ac+b)(bc+a)$$ is wrong.

Try $a=b=\frac{3}{5}.$

Answer 2

Some thoughts.

By Holder, we have \begin{align*} &\left(\sum_{\mathrm{cyc}}\sqrt{\frac{a+b}{c+ab}}\right)^2 \left(\sum_{\mathrm{cyc}} (a + b)^2(c + ab)(c + 2)^3\right)\\[6pt] \ge{}& \left(\sum_{\mathrm{cyc}} (a + b)(c + 2)\right)^3. \end{align*}

It suffices to prove that $$\left(\sum_{\mathrm{cyc}} (a + b)(c + 2)\right)^3 \ge 9 \sum_{\mathrm{cyc}} (a + b)^2(c + ab)(c + 2)^3. \tag{1} $$ (1) is true which is verified by Mathematica - a Computer Algebra System (CAS). It can be proved by the pqr method.

Let $p = a + b + c, q = ab + bc + ca = 3, r = abc$.

(1) is written as $$ \left( 18\,p-189 \right) {r}^{2}+ \left( -216\,{p}^{2}+468\,p+1080 \right) r+64\,{p}^{3}-9\,{p}^{2}-270\,p-1242 \ge 0.$$ Omitted. (Note: Use $q^2 \ge 3pr$ and $p^2 \ge 3q$ and $r \ge \frac{4pq-p^3}{9}$ (degree three Schur).)

Answer 3

Another way.

By AM-GM $$\sum_{cyc}\sqrt{\frac{a+b}{ab+c}}=\sqrt{\sum_{cyc}\left(\frac{a+b}{ab+c}+2\sqrt{\frac{a+b)(a+c}{(c+ab)(b+ac}}\right)}\geq\sqrt{\sum_{cyc}\frac{a+b}{ab+c}+6\sqrt[3]{\prod_{cyc}\frac{a+b}{c+ab}}}$$ and it's enough to prove that: $$\sum_{cyc}\frac{a+b}{ab+c}+6\sqrt[3]{\prod_{cyc}\frac{a+b}{c+ab}}\geq9,$$ which is true by $uvw$, but very ugly.