Solving Inequality Problem 5: xyz=1

[evansmiley]

5. Suppose that x, y and z are positive real numbers such that xyz = 1.
(a) Prove that
27 $\leq$(1 + x + y)$^{2}$ + (1 + y + z)$^{2}$ + (1 + z + x)$^{2}$

with equality if and only if x = y = z = 1.
(b) Prove that
(1 + x + y)$^{2}$ + (1 + y + z)$^{2}$ + (1 + z + x)$^{2}$ $\leq$ 3(x + y + z)$^{2}$

with equality if and only if x = y = z = 1.

The Attempt at a Solution

I don't really how to prove this. I can visualize the truth in my head but I don't know where to start a proof. Does anyone know any particular methods to solve symmetric equalities such as this one? I'm trying to see a nice way to simplify it or introduce a substitution but i can't see anything. Thanks

 

[dikmikkel]

You know that xyz=1 so that would be a good start. And also the sqrt could be taken on each side and then you just subsittute till you have something which you know is greater than $\sqrt{27}$

 

[dibyendu]

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
$$2S_2 \geq 4S_1 + 3 - S_1^2$$

 

[evansmiley]

ok i can see how you rearranged the first inequality but i still don't know where to go from there ? ? ? And may i ask where the inspiration came from to arrange the inequality in that way? Sorry, i don't have much experience solving this kind of problem, so any tips would help immensely! Thanks

 

[dibyendu]

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 more help.

 

[evansmiley]

Oh yes i can see, it comes from the fact that (x-1)^2 ≥ 0
So logically it seems to me to try prove

4x + 4y + 4z + xy + yz +xz ≥ 15

Applying AM-GM we get :

(4x + 4y + 4z) ≥ 12

So now we have

xy + yz + zx ≥ 3

which is equivalent to

1/x + 1/y + 1/z ≥ 3

Now I am stuck :P

 

[dibyendu]

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, $$\frac{x+y+z}{3} \geq 1 \geq \frac{3}{xy + yz + zx}$$

[From the above line, your required proof also follows immediately!]

 

[evansmiley]

Now can you equivalently prove that
$$2S_2 \geq 4S_1 + 3 - S_1^2$$

I tried doing that, however i got 2S_1 rather than 4S_1, did you make an error calculating or was that me?

 

[dibyendu]

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 $$2S_2 \geq 4S_1 + 3 - S_1^2$$

 

[evansmiley]

sorry, yes, you're completely right, i must have made a mistake. Ok so is this logic ok?
2S₂ ≥ 3 + S₁(4-S₁)
Well we already know that S₁≥3 and S₂ ≥ 3
So 2S₂ ≥ 6
On the other hand, if we wish to maximise the right hand side of the equation, we want S₁ to be its minimum, 3, hence the maximum value of the r.h.s. = 3 + 3(4-3) =6
So minimum l.h.s. = 6, max r.h.s. = 6, hence l.h.s. ≥ r.h.s
Equality occurs if x = y = z = 1, because at these values, the min and max, respectively, occur.
Is that rigorous enough a proof? I still didnt manage to prove that equality occurs if and only if x =y =z =1? Thanks

 

[dibyendu]

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 $$LHS = RHS = 6$$

 

[evansmiley]

Ah ok cool, I had tried factoring and leaving a seven outside too before going back and corssing out my work, but I think be both came to the same conclusion. Many thanks for your help dibyendu!

 

[dibyendu]

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