Value of $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}+\frac{1}{\sqrt{1+d^2}}$

Question

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

Answer 1

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.

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

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

3. $x\geq y\geq z\geq1\geq t$,

4. $1\geq x\geq y\geq z\geq t$ and

5. $x\geq y\geq z\geq t\geq1$ for you.

Can you end it now?

Answer 2

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 $$\prod_{cyc}(x^2+3)\geq4((x+y)^2+4)(4+(y+z)^2)\geq$$ $$\geq4(2x+2y+2z+2t)^2=16(x+y+z+t)^2.$$

Answer 3

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: $$(x^2+3)(y^2+3)(z^2+3)(t^2+3)\geq16(x+y+z+t)^2+$$ $$+9\sum_{1\leq i<j\leq4}{(xy-1)^2}+3\sum_{1\leq i<j<k\leq4}{(xyz-1)^2}+(xyzt-1)^2.$$