Inequality

[George Saliaris]

If -π/2 <= a < b < c <= 0 and f(x) = sinx prove that


a [f(b) - f(c) ] + b [f(c) -f(a)] + c[f(a)-f(b)] > 0

My tries
1. Intermediate value Theorem at [a.b] , [b,c] ,[a,c] and use the fact that abs(sinx) <= 1 .
2.I negated the statement and did not reach something.
3 same as 1 but using that abs(sinx) < x.
4. Moving stuff to the right hand side.
5. I proved that a [f(b) - f(c) ] > 0 and c[f(a)-f(b)] > 0 but b [f(c) -f(a)] <0 so I did not go any further.
6. I did the multiplications inside the parenthesis and tried 3,4.
7. I proved that af(b) -bf(a) >0 ,considering g(x) = sinx/x and tried to apply that several times but it did not end up somewhere.



Can somebody please help me with this exercise..I have used over 20 A4 paper and I think I am going crazy...What am I missing?

 

[pka]

If -π/2 <= a < b < c <= 0 and f(x) = sinx prove that
a [f(b) - f(c) ] + b [f(c) -f(a)] + c[f(a)-f(b)] > 0

Look carefully at graph of \(f(x)=\sin(x)\) on \(\left[\frac{-\pi}{2},0\right]\). SEE HERE
The function is increasing, each of \(a,b,c,f(a),f(b),~\&~f(c)\) is negative.
Does that help?

 

[George Saliaris]

Look carefully at graph of \(f(x)=\sin(x)\) on \(\left[\frac{-\pi}{2},0\right]\). SEE HERE
The function is increasing, each of \(a,b,c,f(a),f(b),~\&~f(c)\) is negative.
Does that help?

Not particularly because I also noticed that ..

 

[lex]

The key is that the gradient is increasing there.
Using this fact and the Mean Value Theorem on each of the intervals (a,b) and (b,c), the inequality will drop out quite nicely.

 

[George Saliaris]

If -π/2 <= a < b < c <= 0 and f(x) = sinx prove that


a [f(b) - f(c) ] + b [f(c) -f(a)] + c[f(a)-f(b)] > 0

My tries
1. Intermediate value Theorem at [a.b] , [b,c] ,[a,c] and use the fact that abs(sinx) <= 1 .
2.I negated the statement and did not reach something.
3 same as 1 but using that abs(sinx) < x.
4. Moving stuff to the right hand side.
5. I proved that a [f(b) - f(c) ] > 0 and c[f(a)-f(b)] > 0 but b [f(c) -f(a)] <0 so I did not go any further.
6. I did the multiplications inside the parenthesis and tried 3,4.
7. I proved that af(b) -bf(a) >0 ,considering g(x) = sinx/x and tried to apply that several times but it did not end up somewhere.



Can somebody please help me with this exercise..I have used over 20 A4 paper and I think I am going crazy...What am I missing?

At 1 I meant Mean Value Theorem Not Intermediate Value Theorem (I always get wrong the names of these 2 theorems)

The key is that the gradient is increasing there.
Using this fact and the Mean Value Theorem on each of the intervals (a,b) and (b,c), the inequality will drop out quite nicely.

So I woke up today without checking here and I did exactly what you mentioned and eventually I proved it..