# If $\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}=1$, prove $abcd \geq 3$.

Given four positive real numbers $$a,b,c,d$$. It is given that $$\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}=1$$. Prove $$abcd \geq 3$$.

I've applied AM-HM inequality and got the result, $$a^4+b^4+c^4+d^4 \geq 12$$. Then by AM-GM, $$a^4+b^4+c^4+d^4\geq 4abcd; 12\geq 4abcd \implies 3\geq abcd$$.

But this is contradictory. Then I realized I was doing all wrong. Does anyone have method for this one?

Clearing all denominators, we have that $$\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}=1$$ is equivalent to $$a^4b^4c^4d^4 = (a^4b^4 + a^4c^4 + a^4d^4 + b^4c^4 + b^4d^4 + c^4d^4) + 2(a^4 + b^4 + c^4 + d^4) + 3$$ Applying arithmetic mean $$\ge$$ geometric mean to the brackets on the right gives $$a^4b^4c^4d^4 \ge 6 a^2b^2c^2d^2 + 8 a bcd + 3$$ This is again equivalent to $$(a bcd - 3) (a bcd +1)^3 \ge 0$$ Hence $$a bcd \ge 3$$.

$$\qquad \Box$$

Let $$w=\frac{1}{1+a^4}, x=\frac{1}{1+b^4}$$$$y=\frac{1}{1+c^4}, z=\frac{1}{1+b^4}$$

So, $$a^4=\frac{1-w}{w}=\frac{x+y+z}{w}$$$$b^4=\frac{1-x}{x}=\frac{w+y+z}{x}$$$$c^4=\frac{1-y}{y}=\frac{w+x+z}{y}$$$$d^4=\frac{1-z}{z}=\frac{w+x+y}{z}$$

$$a^4b^4c^4d^4=(\frac{x+y+z}{w}) (\frac{w+y+z}{x})( \frac{w+x+z}{y} )(\frac{w+x+y}{z})-(1)$$

Apply AM-GM for these four terms on RHS individually in eqation $$(1)$$,

RHS$$\geq(\frac{3(xyz)^{\frac{1}{3}}}{w})(\frac{3(wyz)^{\frac{1}{3}}}{x})(\frac{3(wxz)^{\frac{1}{3}}}{y})(\frac{3(wxy)^{\frac{1}{3}}}{z})$$ $$⇒RHS\geq 81$$ RHS is also equal to $$a^4b^4c^4d^4$$.

$$a^4b^4c^4d^4\geq81 ⇒ abcd\geq3$$

• I think it should be 'four terms on RHS' Jan 6, 2023 at 17:34
• Actually there are 4, the 4th one gone down due to plenty of space.
– user1135351
Jan 6, 2023 at 17:38
• I mean that instead on LHS should be RHS? We apply AM-GM to RHS of (1), right? Jan 6, 2023 at 17:39
• Oh yes! Typing mistake. Thank you to made me realize. But the solution correct. Right?
– user1135351
Jan 6, 2023 at 17:40
• Corrected that thing too! Thanks again!
– user1135351
Jan 6, 2023 at 17:43

By AM-GM, $$\frac{a^4}{1+a^4}=\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}\ge \frac{3}{\sqrt[3]{(1+b^4)(1+c^4)(1+d^4)}}$$, etc. Multiplying all these together we have $$\frac{(abcd)^4}{\prod (1+a^4)}\ge \frac{3^4}{\prod (1+a^4)}$$, i.e. $$abcd\ge 3$$.