# Proof for inequality with $a,b,c,d$ with $d =\max(a,b,c,d)$

Let $a,b,c,d$ positive real numbers with $d= \max(a,b,c,d)$. Proof that

$$a(d-c)+b(d-a)+c(d-b)\leq d^2$$

• I believe that the GM-AM inequality with $n=4$ variables might be helpful.

$$\sqrt[n]{x_1 x_2 \dots x_n} \le \frac{x_1+ \dots + x_n}{n}$$

We also know that the Geometric mean is bounded as follows :

$$\min \{x_1, x_2, \dots x_n\} \le \frac{x_1+ \dots + x_n}{n} \le \max \{x_1, x_2, \dots x_n\}$$

** I also tried to draw an square and some rectangles, but nothing worked out.

• Is $d=\max(a,b,c)$ or $d=\max(a,b,c,d)$? – AsdrubalBeltran May 17 '14 at 4:37
• @Arthur Yes, there is. The latter allows $d$ to differ from all of the other three. – Hagen von Eitzen May 17 '14 at 4:44
• according to the question, $d=\max(a,b,c,d)$ – Keith May 17 '14 at 12:56

## 3 Answers

Consider the polynomial $$f(x)=x^3-(a+b+c)x^2+(ab+bc+ac)x-abc$$ having $a,b,c$ as roots. We have $$f(d)=d^3-(a+b+c)d^2+(ab+bc+ac)d-abc=d\cdot(RHS-LHS)-abc$$ and $f(d)=(d-a)(d-b)(d-c)\ge0$.

Or in short $$a(d-c)+b(d-a)+c(d-b)=(a+b+c)d-(ac+ab+bc)\\=\frac{d^3-(d-a)(d-b)(d-c)-abc}{d}\le d^2$$

Divide both sides by $d^2$ to get an equivalent inequality:

$\dfrac{a}{d}\cdot \left(1 - \dfrac{c}{d}\right) + \dfrac{b}{d}\cdot \left(1 - \dfrac{a}{d}\right) + \dfrac{c}{d}\cdot \left(1 - \dfrac{b}{d}\right) \leq 1$.

Now let $x = \dfrac{a}{d}$, $y = \dfrac{b}{d}$, and $z = \dfrac{c}{d}$, then : $0 \leq x, y, z \leq1$, and we are to prove:

$x(1 - z) + y(1 - x) + z(1 - y) \leq 1$.

Consider $f(x,y,z) = x(1 - z) + y(1 - x) + z(1 - y) - 1 = x + y + z - xy - yz - zx - 1$. We find the critical points of $f$. So take partial derivatives:

$f_x = 1 - y - z = 0 \iff y + z = 1$

$f_y = 1 - x - z = 0 \iff x + z = 1$

$f_z = 1 - x - y = 0 \iff x + y = 1$.

Thus $\nabla{f} = 0 \iff x + y = y + z = z + x = 1 \iff x = y = z = \dfrac{1}{2}$. Thus the maximum of $f$ occurs at either the critical values or the boundary points which are: $(x,y,z) = (0,0,0), (0,1,1), ..., (1,1,1)$. Of these values, the max is $0$. So $f(x,y,z) \leq 0$ which is what we are to prove.

• after "Consider" the function $f(x,y,z)$ is defined one way, and right after that another way with $1$ subtracted. [with the subtracted $1$ then one wants to show $f \le 0$ which the rest of your argument does, provided the critical point is shown to be the max]. – coffeemath May 17 '14 at 5:49
• @coffeemath: see edit. – DeepSea May 17 '14 at 5:51
• The critical point is found OK, however the maximum over $0<x,y,z<1$ does not occur there. In fact for example $f(.99,.01,.5)=-.0099>-.25=-1/4.$ – coffeemath May 17 '14 at 6:12
• @coffeemath: agree. This is the local max. The global max attained at the boundary x = y = z = 1 which is 0 – DeepSea May 17 '14 at 6:15
• Yes, the constraints should be $0<x,y,z\le 1$ given the problem statement, and then there are boundary points when one (or more) of $x,y,z$ is $1$. If say $z=1$ then $f(x,y,z)=-xy$ which is less than $0$ [I get $f=-1$ at $x=y=z=1$ however.] I note in your edit you have still allowed $x,y,z$ to be zero, which is excluded in the OP. – coffeemath May 17 '14 at 6:33

The inequality, after multiplying it out and moving all to one side, is $$d^2-ad-bd-cd+ab+ac+bc\ge 0.\tag{1}$$ Since $d=\max(a,b,c,d)$ each of $d-a,\ d-b,\ d-c$ is nonnegative and so $$(d-a)(d-b)(d-c)\ge 0, \\ d^3-ad^2-bd^2-cd^2+abd+acd+bcd\ge abc,$$ where at the last step we moved the $-abc$ term over to the right side. Now since $a,b,c,d$ are positive one can divide both sides of this last inequality by $d$ and obtain $(1)$ as desired, in fact there is the lower bound $abc/d$ for the left side of $(1).$ Note one really only needs $d>0$ (to justify division by $d$) and $a,b,c\ge 0$ for the conclusion to hold.

• In looking at this, it seems really the same idea as in Hagen's answer, without explicitly mentioning the polynomial. I'll delete if anyone thinks it's too similar... – coffeemath May 17 '14 at 11:25