Inequality

First, we know that $a^2+b^2+c^2+d^2\ge ab+bc+cd+da$ $\displaystyle\iff (a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2\ge 0$

We'll use this fact in our proof.

$\displaystyle \frac{a^3+c^3}{2\sqrt{bd}}+\frac{b^3+d^3}{2\sqrt{ac}}$ $\displaystyle \stackrel{\text{AM-GM}}\ge\frac{a^3+c^3}{b+d}+\frac{b^3+d^3}{a+c}$ $=\frac{a^3}{b+d}+\frac{b^3}{a+c}+\frac{c^3}{b+d}+\frac{d^3}{a+c}$ $\displaystyle =\frac{a^4}{ab+da}+\frac{b^4}{ab+ac}+\frac{c^4}{bc+cd}+\frac{d^4}{da+cd}$ $\displaystyle\stackrel{\text{Cauchy-Schwarz}}\ge \frac{(a^2+b^2+c^2+d^2)^2}{2(ab+bc+cd+da)}$ $\displaystyle\ge \frac{(ab+bc+cd+da)^2}{2(ab+bc+cd+da)}= \frac{1}{2}$

The only possible case of equality in $a^2+b^2+c^2+d^2\ge ab+bc+cd+da$ is when $a=b=c=d$ .

If $a=b=c=d$ , then $\displaystyle \frac{a^3+c^3}{b+d}+\frac{b^3+d^3}{a+c}=2\cdot\frac{2a^3}{2a}=2a^2$

And this is equal to $\frac{1}{2}$ if and only if $a=\frac{1}{2}$ .

Hence, $k=\frac{1}{2}$ and equality occurs if and only if $a_{\text{eq}}=b_{\text{eq}}=c_{\text{eq}}=d_{\text{eq}}=\frac{1}{2}$ , all adding up to $\boxed{2.5}$ .

This problem is original, but I thought of it while solving this . They are still very different problems with different methods, though.

(a^3+c^3)/2 >=(ac)^3/2 (AM-GM),similiar with b and d

That means that (a^3+c^3)/(2(bd)^0.5)+(b^3+d^3)/(2(ac)^0.5)>=

(ac)^1.5/(bd)^0.5+(bd)^1.5/(ac)=((ac)^2+(bd)^2)/(abcd)^0.5>=2(abcd)^0.5

1/4= (ab+bc+cd+da)/4>abcd (AM-GM) so abcd is max 0.0625 when a=b=c=d=0.5 so our inequality is

=2*0.0625^0.5=0.5

Equality: a=c,b=d for first use of AM-GM,ac=bd for second and a=b=c=d for last

That means a=b=c=d=0.5=k, so a+b+c+d+k=2.500