7
$\begingroup$

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?

$\endgroup$

3 Answers 3

9
$\begingroup$

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$.

$\qquad \Box$

$\endgroup$
0
6
$\begingroup$

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 $a^4b^4c^4d^4$.

$$a^4b^4c^4d^4\geq81 ⇒ abcd\geq3$$

$\endgroup$
6
  • $\begingroup$ I think it should be 'four terms on RHS' $\endgroup$
    – mowzorn
    Jan 6, 2023 at 17:34
  • 1
    $\begingroup$ Actually there are 4, the 4th one gone down due to plenty of space. $\endgroup$
    – user1135351
    Jan 6, 2023 at 17:38
  • $\begingroup$ I mean that instead on LHS should be RHS? We apply AM-GM to RHS of (1), right? $\endgroup$
    – mowzorn
    Jan 6, 2023 at 17:39
  • 1
    $\begingroup$ Oh yes! Typing mistake. Thank you to made me realize. But the solution correct. Right? $\endgroup$
    – user1135351
    Jan 6, 2023 at 17:40
  • 2
    $\begingroup$ Corrected that thing too! Thanks again! $\endgroup$
    – user1135351
    Jan 6, 2023 at 17:43
1
$\begingroup$

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. $abcd\ge 3$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .