Another Inequality

We will prove the inequality |a-c|(√1+a²)(√1+c²)+ |b-c|√(1+b²)√(1+c²)≥|a-b|√(1+a²)√(1+b²) is true.(a,b,c are real)

That doesn't remind of any direct inequality. Tried simplification ,it became more complex. the inequality is not symmetric , not homogenous ,neither direct modulus inequality. Tried using trigonometric replacements. It's quite natural to take a = tanA , b = tanB , c= tanC (all the angles A,B,C lies between -pi/2 and pi/2)

The inequality boils down to |sin(A-C)| + |sin(C-B)|≥|sin(A-B)|=| sin{(A-C) + (C-B)}| =|sin(A-C)cos(C-B) + sin(C-B)cos(A-C)|

we have |sin(A-C)|+ |sin(C-B)| ≥|sin(A-c)||cos(C-B)| + |sin(C-B)||cos(A-C)|

(Note that |cosx| lies between 0 to 1)

≥|sin(A-C)cos(C-B) + sin(C-B)cos(A-C)| (by modulus inequality)

though Ricky did nt say anything,the solution is correct according to me..

Great work yet again Shubhodip , nice :) .