Recent content by dibyendu

You are most welcome, Evansmiley. It's my pleasure.

Observe that 4S_1 + 3 - S_1^2 = 7 - (S_1 - 2)^2 Since S_1 \geq 3 the RHS is always \leq 6 Finally the LHS is always \geq 6 and the RHS is always \leq 6 , which completes the proof. You can solve the equality case by considering the equations LHS = RHS = 6

Please check my calculations. I did it in a hurry. We are required TPT (S_1 - x + 1)^2 + (S_1 - y + 1)^2 + (S_1 - z + 1)^2 \leq 3S_1^2 Simplifying, we find that we are required TPT S_1^2 - 2S_2 \leq 2S_1^2 - 4S_1 - 3 i.e., TPT 2S_2 \geq 4S_1 + 3 - S_1^2

A simpler way to complete the proof is to use the following facts: x + y + z \geq 3 and xy + yz + zx \geq 3 These facts follow from the AM-GM-HM inequality: \frac{x+y+z}{3} \geq (xyz)^{\frac{1}{3}} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} Which means...

The idea is to apply the AM-GM-HM inequality at some stage or the other. If you keep that in mind, you can try to manipulate the expressions accordingly. Can you see that (x+1)^2 \geq 4x (y+1)^2 \geq 4y (z+1)^2 \geq 4z Can you try to complete the proof now? Let me know if you need...

Hints: 5. (a) Rewriting the expressions a bit, can you equivalently prove that (x+1)^2 + (y+1)^2 + (z+1)^2 + xy + yz + zx \geq 15 5. (b) Put x + y + z = S_1 xy + yz + zx = S_2 and rewrite the expressions. Now can you equivalently prove that 2S_2 \geq 4S_1 + 3 - S_1^2