<s>[INST] Value of $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}+\frac{1}{\sqrt{1+d^2}}$ 
If  $$\frac{a+b+c+d}{\sqrt{(1+a^2)(1+b^2)(1+c^2)(1+d^2)}}= \frac{3\sqrt{3}}{4}$$ for $a,b,c,d>0$
Then
Value of $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}+\frac{1}{\sqrt{1+d^2}}$ is

Try: using A.M G.M Inequality
$$1+a^2\geq 2a\;, 1+b^2\geq 2b\;,1+c^2\geq 2c\; 1+d^2\geq 2d$$
$$\sqrt{(1+a^2)(1+b^2)(1+c^2)(1+d^2)}\geq 4\sqrt{abcd}$$
$$\frac{a+b+c+d}{\sqrt{(1+a^2)(1+b^2)(1+c^2)(1+d^2)}}\leq \frac{a+b+c+d}{4\sqrt{abcd}}$$
I have edited it, please have a look
Could some help me to solve it , Thanks
 [/INST] With thanks to Michael for pointing this beautiful topic.
For proving
$$(x^2+3)(y^2+3)(z^2+3)(t^2+3)\geq16(x+y+z+t)^2,$$
I have used the following result discovered by me:
$$(x^2+3)(y^2+3)(z^2+3)(t^2+3)\geq16(x+y+z+t)^2+$$
$$+9\sum_{1\leq i<j\leq4}{(xy-1)^2}+3\sum_{1\leq i<j<k\leq4}{(xyz-1)^2}+(xyzt-1)^2.$$
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