SOLUTION TO PROBLEM 9

Let

$$x:= b+c-a, \qquad y:= c+a-b, \qquad z:= a+b-c.$$

Observe that x>0, y>0, z>0 since any side of the triangle is smaller than the sum of the remaining two sides.

We solve the system for a, b, and c and get

$$a = {{y+z}\over 2}, \qquad b = {{x+z}\over 2}, \qquad c = {{x+y}\over 2}.$$

Thus we can rewrite the initial inequality in the following equivalent form

$${{y+z}\over 2x} + {{x+z}\over 2y} + {{x+y}\over 2z} \ge 3,$$

and therefore it is enough to prove this last inequality. Now observe that

$${{y+z}\over 2x} + {{x+z}\over 2y} + {{x+y}\over 2z} = {1 \ove... ...z\over x}+{x\over z}) + ({z\over y}+{y\over z})]. \leqno {(1)}$$

Note that in general

$${m\over n} + {n\over m} \ge 2, \qquad m, n > 0,$$

since this is equivalent with $(m-n)^2 \ge 0$ which is true. Applying this inequality to all the terms in (1), we obtain

$${{y+z}\over 2x} + {{x+z}\over 2y} + {{x+y}\over 2z} \ge {1\over 2} [2 + 2 + 2] = 3.$$

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Fri Oct 26 13:11:29 CDT 2001