1. ## interesting inequality

If a,A,b,B,c,C are non-negative real numbers, where
a+A=b+B=c+C=s,
than
a.B+b.C+c.A < s^2

Can you help me prove this inequality only with using properties of linear function?

2. ## interesting inequality

What if A,b,B,c and C all equal 0? Then I believe a=S for all values of a, and the inequality does not hold (if I'm interpreting the formula correctly - I'm assuming, for example, that a.B would be 1.0 if a=1 and B=0).

- Steve J

3. Originally Posted by Steve_J
What if A,b,B,c and C all equal 0?
Then $\displaystyle s=0.$ The inequality is strict if $\displaystyle s>0.$

4. You can rewrite the equation
a.B + b.C + c.A as

$\displaystyle (a + 1/10 B) + (b + 1/10 C) + (c + 1/10 A)$ assuming that the positive numbers are integers and only hold one place value. A rational fraction or an irrational number wouldn't make sense if you mean to use ' . ' as a decimal point

Follows that this is equivalent to
$\displaystyle 1/10(A + B + C) + (a + b + c)$
$\displaystyle 1/10 [(A + B +C) + 10(a + b + c)]$

Try solving from here.

5. Originally Posted by mptlu6
If a,A,b,B,c,C are non-negative real numbers, where
a+A=b+B=c+C=s,
than
a.B+b.C+c.A < s^2

Can you help me prove this inequality only with using properties of linear function?
I'm not sure what you mean by "only with using properties of linear function". But here is a proof that uses only elementary algebra.

Step 1. If x ≤ 1 and y ≤ 1 then x + y ≤ 1 + xy.
Proof. Follows from the fact that (1 – x)(1 – y) ≥ 0.

Step 2. If –1 ≤ x ≤ 1, –1 ≤ y ≤ 1 and –1 ≤ z ≤ 1, then yz + zx + xy + 1 ≥0.
Proof. The result clearly holds if x, y and z are all positive, or all negative. (It also holds if any of them are zero.) So suppose that x and y have one sign, and z has the opposite sign. In fact, changing the sign of all three variables (which does not change the value of yz + zx + xy + 1), we may assume that x and y are positive and that z is negative, say z = –w, with 0 ≤ w ≤ 1. Then yz + zx + xy + 1 = 1 + xy – w(x+y) ≥ 1 + xy – (x+y) ≥ 0, by step 1.

Step 3. Now suppose that a, A, b, B, c, C are as in the statement of the problem. Since A = s–a is nonnegative, as is a, it follows that
0 ≤ a ≤ s, and similarly for b and c. Therefore $\displaystyle \frac{2a-s}s$ lies between –1 and 1, with similar results for b and c.

Step 4. Now apply step 2, with $\displaystyle x = \frac{2a-s}s,\ y = \frac{2b-s}s,\ z = \frac{2c-s}s$. It follows that $\displaystyle (2a-s)(2c-s) + (2c-s)(2a-s) + (2a-s)(2b-s) + s^2 \geqslant0$. Multiply out the brackets and simplify this, to get $\displaystyle s(a+b+c) - (bc+ca+ab)\leqslant s^2$.

Step 5. The original problem was to find an upper bound for aB + bC + cA. But this is equal to a(s–b) + b(s–c) + c(s–a) = s(a+b+c) – (bc+ca+ab). So the result follows from step 4.

[The inequality need not be strict. It becomes an equality, for example, if a=b=0 and c=s.]

6. Originally Posted by Opalg
Step 3. Now suppose that a, A, b, B, c, C are as in the statement of the problem. Since A = s–a is nonnegative, as is a, it follows that
0 ≤ a ≤ s, and similarly for b and c. Therefore $\displaystyle \frac{2s-a}s$ lies between –1 and 1, with similar results for b and c.
I dont get this step..

$\displaystyle a,A \geq 0 \implies s \geq 0$. Thus $\displaystyle \frac{2s - a}{s} \leq 1 \Leftrightarrow 2s - a \leq s \Leftrightarrow s - a = A \leq 0$

But thats not true....

7. Originally Posted by Isomorphism
I dont get this step..

$\displaystyle a,A \geq 0 \implies s \geq 0$. Thus $\displaystyle \frac{2s - a}{s} \leq 1 \Leftrightarrow 2s - a \leq s \Leftrightarrow s - a = A \leq 0$

But thats not true....
Stupid mistake on my part. I have corrected it to read 2a–s instead of 2s–a.