Prove this inequality $\sum_{cyc}\sqrt[4]{\frac{a}{b+c}}\ge \sqrt[4]{16+\frac{196abc}{(a+b)(b+c)(c+a)}}$ Let $a;b;c\ge0$. Prove the inequality $$\sqrt[4]{\dfrac{a}{b+c}}+\sqrt[4]{\dfrac{b}{c+a}}+\sqrt[4]{\dfrac{c}{a+b}}\ge\sqrt[4]{16+\dfrac{196abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}$$

I think we should let $\frac{a}{b+c};\frac{b}{c+a};\frac{c}{a+b}\rightarrow x;y;z$ and use Schur's ineq but i don't know how to annul the radicals and the inequality isn't homogeneous so i can't solve it.
 A: We need to prove that
$$\left(\sum_{cyc}\sqrt[4]{a(a+b)(a+c)}\right)^4\geq16\prod_{cyc}(a+b)+196abc.$$
Now, by Holder, C-S,AM-GM and Schur we obtain:
$$\left(\sum_{cyc}\sqrt[4]{a(a+b)(a+c)}\right)^4=\sum_{cyc}a(a+b)(a+c)+$$
$$+4\sum_{cyc}\sqrt[4]{(a^2(a+b+c)+abc)^3(b^2(a+b+c)+abc)}+$$
$$+4\sum_{cyc}\sqrt[4]{(a^2(a+b+c)+abc)^3(c^2(a+b+c)+abc)}+$$
$$+6\sum_{cyc}\sqrt{(a^2(a+b+c)+abc)(b^2(a+b+c)+abc)}+$$
$$+12\sum_{cyc}\sqrt[4]{a^2bc(a+b)^3(a+c)^3(b+c)^2}\geq$$
$$\geq\sum_{cyc}(a^3+a^2b+a^2c+abc)+4\sum_{cyc}\left(\left(\sqrt{a^3b}+\sqrt{a^3c}\right)(a+b+c)+2abc\right)+$$
$$+6\sum_{cyc}(ab(a+b+c)+abc)+144abc\geq$$
$$\geq\sum_{cyc}\left(a^3+7a^2b+7a^2c+4\sqrt{a^5b}+4\sqrt{a^5c}+8\sqrt{a^3b^3}+77abc\right)\geq$$
$$\geq\sum_{cyc}\left(8a^2b+8a^2c+4\sqrt{a^5b}+4\sqrt{a^5c}+8\sqrt{a^3b^3}+76abc\right)$$ and since
$$16\prod_{cyc}(a+b)+196abc=\sum_{cyc}(16^2b+16a^2c+76abc)\geq0,$$ it's enough to prove that
$$\sum_{cyc}\left(4\sqrt{a^5b}+4\sqrt{a^5c}-8a^2b-8a^2c+8\sqrt{a^3b^3}\right)\geq0$$ or
$$\sum_{cyc}\sqrt{ab}\left(a^2-2\sqrt{a^3b}+2ab-2\sqrt{ab^3}+b^2\right)\geq0$$ or
$$\sum_{cyc}\sqrt{ab}(a+b)(\sqrt{a}-\sqrt{b})^2\geq0.$$
Done!
