<s>[INST] Find the maximum value of the expression Let a,b,c,d be positive real numbers. Show that

$$\frac{(ab+bc+ca)}{(a^3+b^3+c^3)}+\frac{(ab+bd+da)}{(a^3+b^3+d^3)}+\frac{(ac+cd+da)}{(a^3+c^3+d^3)}+\frac{ (bc+cd+db)}{(b^3+c^3+d^3)} \le \frac{(a^2+b^2)}{(ab)^{3/2}} + \frac{(c^2+d^2)}{(cd)^{3/2}} +\frac{(a^2+c^2)}{(ac)^{3/2}} + \frac{(b^2+d^2)}{(bd)^{3/2}} +\frac{ (a^2+d^2)}{(ad)^{3/2}} + \frac{(b^2+c^2)}{(bc)^{3/2}}$$ 

 [/INST] Note that $\displaystyle \frac{a^2+b^2}{(ab)^{3/2}} \ge 2\frac{a^2+b^2}{a^3+b^3}$ by AM-GM.  Using that, we get
$$\frac{a^2+b^2}{(ab)^{3/2}}+\frac{b^2+c^2}{(b c)^{3/2}}+\frac{c^2+a^2}{(c a)^{3/2}} \ge  2\left(\frac{a^2+b^2}{a^3+b^3} + \frac{b^2+c^2}{b^3+c^3} + \frac{c^2+a^2}{c^3+a^3} \right) \tag{1}$$
Using Cauchy-Schwarz inequality on the RHS of $(1)$ above, we have
$$\frac{a^2+b^2}{(ab)^{3/2}}+\frac{b^2+c^2}{(b c)^{3/2}}+\frac{c^2+a^2}{(c a)^{3/2}} \ge 2\frac{(a+b+c)^2}{a^3+b^3+c^3} \ge 6\frac{ab+bc+ca}{a^3+b^3+c^3} \tag{2}$$
Now cyclically summing four similar expressions using $(2)$, we get
$$\sum_{sym} \frac{a^2+b^2}{(ab)^{3/2}} \ge 2\sum_{sym} \frac{ab+bc+ca}{a^3+b^3+c^3}$$
which is tighter than the inequality in your problem...
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