By Cauchy-Schwarz or Jensen, we have $$\sum_{cyc}{\sqrt{\dfrac{a}{b+c+d^2+a^3}}}\leq \sqrt{4\sum_{cyc}{\dfrac{a}{b+c+d^2+a^3}}}\leq 2$$It sufficies to show $$\sum_{cyc}{\dfrac{a}{b+c+d^2+a^3}}\leq 1$$But by C-S $$\left(b+c+d^2+a^3\right)\left(b+c+1+\dfrac{1}{a}\right)\geq \left(a+b+c+d\right)^2$$then $$\sum_{cyc}{\dfrac{a}{b+c+d^2+a^3}}\leq \dfrac{\sum\limits_{cyc}{a\left(b+c+1+\dfrac{1}{a}\right)}}{\left(a+b+c+d\right)^2}$$$$=\dfrac{\sum\limits_{sym}{(ab)}+ac+bd+a+b+c+d+4}{\left(a+b+c+d\right)^2}\overbrace{\leq}^{?} 1$$which is easy since $$a^2+b^2+c^2+d^2+\left(a+c\right)\left(b+d\right)\geq a+b+c+d+4$$$$\Longleftrightarrow a^2+b^2+c^2+d^2\geq a+b+c+d$$where the last inequality is true by $abcd=1$. .

I think that main idea in the problem is the same with JBMO 2019 #A.5. These problems are very related in their basis.

ehuseyinyigit wrote: By Cauchy-Schwarz or Jensen, we have $$\sum_{cyc}{\sqrt{\dfrac{a}{b+c+d^2+a^3}}}\leq \sqrt{4\sum_{cyc}{\dfrac{a}{b+c+d^2+a^3}}}\leq 2$$It sufficies to show $$\sum_{cyc}{\dfrac{a}{b+c+d^2+a^3}}\leq 1$$But by C-S $$\left(b+c+d^2+a^3\right)\left(b+c+1+\dfrac{1}{a}\right)\geq \left(a+b+c+d\right)^2$$then $$\sum_{cyc}{\dfrac{a}{b+c+d^2+a^3}}\leq \dfrac{\sum\limits_{cyc}{a\left(b+c+1+\dfrac{1}{a}\right)}}{\left(a+b+c+d\right)^2}$$$$=\dfrac{\sum\limits_{sym}{(ab)}+ac+bd+a+b+c+d+4}{\left(a+b+c+d\right)^2}\overbrace{\leq}^{?} 1$$which is easy since $$a^2+b^2+c^2+d^2+\left(a+c\right)\left(b+d\right)\geq a+b+c+d+4$$$$\Longleftrightarrow a^2+b^2+c^2+d^2\geq a+b+c+d$$where the last inequality is true by $abcd=1$. . can you expl why a^2 + b^2 + c^2 + d^2 >= a + b + c + d for abcd = 1

Duk168 wrote: can you expl why a^2 + b^2 + c^2 + d^2 >= a + b + c + d for abcd = 1 It's Muirhead

Duk168 wrote: ehuseyinyigit wrote: By Cauchy-Schwarz or Jensen, we have $$\sum_{cyc}{\sqrt{\dfrac{a}{b+c+d^2+a^3}}}\leq \sqrt{4\sum_{cyc}{\dfrac{a}{b+c+d^2+a^3}}}\leq 2$$It sufficies to show $$\sum_{cyc}{\dfrac{a}{b+c+d^2+a^3}}\leq 1$$But by C-S $$\left(b+c+d^2+a^3\right)\left(b+c+1+\dfrac{1}{a}\right)\geq \left(a+b+c+d\right)^2$$then $$\sum_{cyc}{\dfrac{a}{b+c+d^2+a^3}}\leq \dfrac{\sum\limits_{cyc}{a\left(b+c+1+\dfrac{1}{a}\right)}}{\left(a+b+c+d\right)^2}$$$$=\dfrac{\sum\limits_{sym}{(ab)}+ac+bd+a+b+c+d+4}{\left(a+b+c+d\right)^2}\overbrace{\leq}^{?} 1$$which is easy since $$a^2+b^2+c^2+d^2+\left(a+c\right)\left(b+d\right)\geq a+b+c+d+4$$$$\Longleftrightarrow a^2+b^2+c^2+d^2\geq a+b+c+d$$where the last inequality is true by $abcd=1$. . can you expl why a^2 + b^2 + c^2 + d^2 >= a + b + c + d for abcd = 1 u can also us c.s like; $(1+1+1+1)(a^2+b^2+c^2+d^2) \ge (a+b+c+d)^2$ $a^2+b^2+c^2+d^2 \ge \frac{(a+b+c+d)^2}{4}$ $(a+b+c+d)^2 \ge 4*(a+b+c+d)$ ==> $a+b+c+d \ge 4$ which is true by $AM-GM$