For positive real numbers a, b and c, find the minimum value of P = (a/(a+b))^4 + (b/(b+c))^4 + (c/(c+a))^4.

Step-by-step solution

1. 1

   Change variables to expose the constraint. Put $u=\dfrac{b}{a}$, $v=\dfrac{c}{b}$, $w=\dfrac{a}{c}$, so that $uvw=1$ automatically. Then $x=\frac{a}{a+b}=\frac{1}{1+u},\quad y=\frac{1}{1+v},\quad z=\frac{1}{1+w},\qquad P=x^4+y^4+z^4.$

2. 2

   Guess the equality case. Taking $a=b=c$ gives $x=y=z=\tfrac12$ and $P=3\left(\frac12\right)^4=\frac{3}{16}.$ The rest of the work is proving no configuration beats this.

3. 3

   Step down through the sum of squares. By the power-mean inequality applied to $x^2,y^2,z^2$, $x^4+y^4+z^4\ge\frac{\left(x^2+y^2+z^2\right)^2}{3}.$ So it suffices to show $x^2+y^2+z^2\ge\tfrac34$. (The tempting shortcut $x+y+z\ge\tfrac32$ is simply false — $a=1,b=2,c=4$ gives $1.4667$ — so the squares, not the first powers, are the right quantity to bound.)

4. 4

   Use a two-variable lemma. For all $u,v>0$, $\frac{1}{(1+u)^2}+\frac{1}{(1+v)^2}\ge\frac{1}{1+uv},$ because clearing denominators leaves the identity $(uv-1)^2+uv(u-v)^2\ge 0$.

5. 5

   Apply it with $uv=1/w$. The lemma gives $x^2+y^2\ge\dfrac{1}{1+1/w}=\dfrac{w}{1+w}=1-z$, so $x^2+y^2+z^2\ge 1-z+z^2=\left(z-\frac12\right)^2+\frac34\ge\frac34.$

6. 6

   Combine and confirm. Therefore $P\ge\frac{(3/4)^2}{3}=\frac{3}{16},$ with equality exactly when $z=\tfrac12$ and $u=v$, i.e. $a=b=c$. A numerical global search over $200$ random restarts also bottoms out at $0.1875$ with $a:b:c=1:1:1$.