cooked proofs q (2 Viewers)

[C2H6O]

Any ideas?

 

[cheesynooby]

cooked

 

[Sylfiphy]

Any ideas?

surely this is AM-GM right

 

[killer queen]

surely this is AM-GM right

we haven't even done AM/GM uh oh

 

[v.tex]

LHS-RHS expand prove greater than equal to zero somehow idk

otherwise will be am-gm

 

[C2H6O]

I’ll be honest I never formally learned wtf am-gm means, but does that mean you use some form of ?

 

[v.tex]

I’ll be honest I never formally learned wtf am-gm means, but does that mean you use some form of ?

bro u e4ed 4u and didnt know the Am-Gm?

 

[C_master]

I’ll be honest I never formally learned wtf am-gm means, but does that mean you use some form of ?

didn't u accelerate 3u 4u already or is that cap

 

[killer queen]

LHS-RHS expand prove greater than equal to zero somehow idk

otherwise will be am-gm

I tried LHS-RHS and I might be slow but I couldn't figure out any manipulation to get there, am I tripping

 

[v.tex]

I tried LHS-RHS and I might be slow but I couldn't figure out any manipulation to get there, am I tripping

yeah there isnt any its am gm

 

[C2H6O]

didn't u accelerate 3u 4u already or is that cap

I lowkey thought am-gm was oos I only heard of it from bos after doing the hsc. (I ain’t gonna bother defending the allegations anymore I realise how ridiculous it is that i actually passed mx2)

 

[killer queen]

I lowkey thought am-gm was oos I only heard of it from bos after doing the hsc. (I ain’t gonna bother defending the allegations anymore I realise how ridiculous it is that i actually passed mx2)

Wait if you already did it then why the question?

Also if anyone figures it out please post working

 

[C2H6O]

Wait if you already did it then why the question?

Also if anyone figures it out please post working

touche. idk ig im just curious how you'd solve it

 

[cheesynooby]

touche. idk ig im just curious how you'd solve it

im curious how u did it now

 

[C2H6O]

im curious how u did it now

I haven’t solved the q that’s why I posted it

 

[ButteryNucleotide]

ChatGPT used AM-GM inequality:

We wish to show that for all x, y > 1 the following inequality holds:
  x⁵ + y⁵ ≥ xy · (x² + y²).

The proof proceeds in several clear steps.

─────────────────────────────
Step 1. Apply AM–GM to x⁵ and y⁵

For any two positive numbers, the Arithmetic Mean–Geometric Mean (AM–GM) inequality gives

  (x⁵ + y⁵) / 2 ≥ √(x⁵y⁵).

Multiplying both sides by 2 yields

  x⁵ + y⁵ ≥ 2·√(x⁵y⁵).

Since √(x⁵y⁵) = (xy)^(5/2), we have

  x⁵ + y⁵ ≥ 2·(xy)^(5/2).

─────────────────────────────
Step 2. Find a Lower Bound for xy · (x² + y²)

Again, applying AM–GM to x² and y² we obtain

  (x² + y²) / 2 ≥ √(x² · y²) = xy.

Multiplying by 2 gives

  x² + y² ≥ 2xy.

Now, multiplying by xy (which is positive since x, y > 1) leads to

  xy · (x² + y²) ≥ 2x y · (xy) = 2·(xy)².

─────────────────────────────
Step 3. Compare the Two Results

We now have two inequalities:
  (1) x⁵ + y⁵ ≥ 2·(xy)^(5/2),
  (2) xy (x² + y²) ≥ 2·(xy)².

To prove the desired inequality x⁵ + y⁵ ≥ xy (x² + y²), it suffices to show that

  2·(xy)^(5/2) ≥ 2·(xy)².

Since 2 is common to both sides, we need only prove

  (xy)^(5/2) ≥ (xy)².

Because x, y > 1, the product xy > 1. Dividing both sides by (xy)² (valid because (xy)² > 0) gives

  (xy)^(5/2 - 2) = (xy)^(1/2) ≥ 1.

But since xy > 1, raising xy to the power ½ yields a number greater than 1. Hence, (xy)^(1/2) ≥ 1 is indeed true. This confirms

  (xy)^(5/2) ≥ (xy)².

─────────────────────────────
Step 4. Conclude the Proof

Chaining the inequalities together, we have:

  x⁵ + y⁵ ≥ 2·(xy)^(5/2)  [by Step 1]
       ≥ 2·(xy)²      [by Step 3]
       ≤ xy · (x² + y²)   [by Step 2].

Thus, the inequality

  x⁵ + y⁵ ≥ xy · (x² + y²)

holds true for all x, y > 1.

─────────────────────────────
Equality Condition

Notice that equality in the AM–GM inequality occurs if and only if the two numbers are equal. For Step 1 we require x⁵ = y⁵ which means x = y, and for Step 2 we require x² = y², again giving x = y. Finally, for the reduction (xy)^(1/2) = 1 we must have xy = 1. With x = y, this forces x² = 1 and hence x = 1. Therefore, equality is achieved if and only if x = y = 1.

Since our proof is for x, y > 1, while the equality case occurs at the boundary x = y = 1, the inequality is strict in the domain x, y > 1.

─────────────────────────────
Conclusion

For all x, y > 1, we have demonstrated through a series of applications of the AM–GM inequality that

  x⁵ + y⁵ ≥ xy · (x² + y²).

This constitutes a complete, step-by-step, holistic proof in the style of HSC Extension 2 mathematics.

 

[spikechook]

 

[v.tex]

oh hell nah

 

[v.tex]

If you can't do this ur cooked

 

[killer queen]

View attachment 46912If you can't do this ur cooked

is that as simple as I think it is