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
 A: We'll prove that
$$\frac{a+b+c+d}{\sqrt{(1+a^2)(1+b^2)(1+c^2)(1+d^2)}}\leq\frac{3\sqrt3}{4},$$ where the equality occurs for $a=b=c=d=\frac{1}{\sqrt3}$ only.
Indeed, let $a=\frac{x}{\sqrt3},$ $b=\frac{y}{\sqrt3},$ $c=\frac{z}{\sqrt3}$ and $d=\frac{t}{\sqrt3}.$
Thus, we need to prove that
$$(x^2+3)(y^2+3)(z^2+3)(t^2+3)\geq16(x+y+z+t)^2.$$
Consider five cases.


*

*$x\geq1\geq y\geq z\geq t$.


Thus, by C-S $$\prod_{cyc}(x^2+3)=(x^2+3)\prod_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1+4)=$$
$$=(x^2+3)\left(4^3+4^2\sum_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1)+4\sum_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1)(z^2-1)+\prod_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1)\right)=$$
$$=(x^2+3)\left(4^3+4^2\sum_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1)+4\sum_{y\rightarrow z\rightarrow t\rightarrow y}(1-y^2)(1-z^2)-\prod_{y\rightarrow z\rightarrow t\rightarrow y}(1-y^2)\right)\geq$$
$$\geq(x^2+3)\left(4^3+4^2\sum_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1)+4\sum_{y\rightarrow z\rightarrow t\rightarrow y}(1-y^2)(1-z^2)-(1-y^2)(1-z^2)\right)\geq$$
$$\geq(x^2+3)\left(4^3+4^2\sum_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1)\right)=16(x^2+1+1+1)(1+y^2+z^2+t^2)\geq$$
$$\geq16(x+y+z+t)^2;$$
2. $x\geq y\geq1\geq z\geq t.$
Thus, by C-S again we obtain:
$$\prod_{cyc}(x^2+3)=\prod_{cyc}(4+(x^2-1))\geq(16+4(x^2+y^2-2))(16+4(z^2+t^2-2))=$$
$$=16(x^2+y^2+1+1)(1+1+z^2+t^2)\geq16(x+y+z+t)^2;$$
The cases 


*$x\geq y\geq z\geq1\geq t$,

*$1\geq x\geq y\geq z\geq t$ and

*$x\geq y\geq z\geq t\geq1$ for you.
Can you end it now?
A: Another way for the proof of the inequality $(x^2+3)(y^2+3)(z^2+3)(t^2+3)\geq16(x+y+z+t)^2.$
We obtain: $$(x^2+3)(y^2+3)=x^2y^2+3(x^2+y^2)+9=$$
$$=(xy-1)^2+(x-y)^2+2(x+y)^2+8\geq2((x+y)^2+4).$$
By the same way $$(z^2+3)(t^2+3)\geq2((z+t)^2+4).$$
Id est, by C-S $$\prod_{cyc}(x^2+3)\geq4((x+y)^2+4)(4+(y+z)^2)\geq$$
$$\geq4(2x+2y+2z+2t)^2=16(x+y+z+t)^2.$$
A: 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:
$$(x^2+3)(y^2+3)(z^2+3)(t^2+3)\geq16(x+y+z+t)^2+$$
$$+9\sum_{1\leq i<j\leq4}{(xy-1)^2}+3\sum_{1\leq i<j<k\leq4}{(xyz-1)^2}+(xyzt-1)^2.$$
