# Incels.is - Involuntary Celibate

##### Welcome! This is a forum for involuntary celibates: people who lack a significant other. Are you lonely and wish you had someone in your life? You're not alone! Join our forum and talk to people just like you. #### uranium235

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I'm not convinced this is true. Take P(x) = ¼ + x. In that case your desideratum amounts to ¼ + ab ≤ (¼ + a)(¼ + b) for positive a & b, but this is evidently false for a = b = ¼.
Sorry my mistake it was Sqrt(p(a)p(b))=p(Sqrt(ab)) #### Ahnfeltia

##### 3254011055738
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Sorry my mistake it was Sqrt(p(a)p(b))=p(Sqrt(ab))
The equality sign should have been a nonstrict inequality. Anyhoo, this reminds me of midpoint (or Jensen) convexity, albeit with the geometric mean instead of the arithmetic one.

Let g = sqrt(ab). We want to prove that p(g)p(g) ≤ p(a)p(b). Let p(x) = c_0 + c_1*x + ... + c_n*x^n. Upon expanding the product p(a)p(b) we get a sum of terms of the form c_i*c_k*a^i*b^k. If we can therefore manage to show that c_i^2*g^(2i) ≤ c_i^2*a^i*b^i & c_i*c_k*2g^(i+k) ≤ c_i*c_k*(a^i*b^k + a^k*b^i) for every i ≠ k, we are done (since a & b & the c_i are all strictly positive). The former inequality is obviously an equality and therefore true. Since the c_i are all strictly positive, the latter inequality is tantamount to showing that 2g^(i+k) ≤ a^i*b^k + a^k*b^i.

Showing that 2g^(i+k) ≤ a^i*b^k + a^k*b^i is easy with Lagrange multipliers. For concreteness, I'll illustrate the case where i = 5 & k = 2. In that case, we want to show that 2g^7 ≤ a^5*b^2 + a^2*b^5 = a^2*b^2*(a^3 + b^3) = g^4*(a^3 + b^3) -- i.e., 2g^3 ≤ a^3 + b^3. Now we use Lagrange multipliers. We want to minimize the bivariate function x^3 + y^3 constrained to the hyperbola xy = g^2. Lagrange multipliers readily yields that 3x^2/y = 3y^2/x -- i.e., x^3 = y^3, so x = y = g (assuming x & y to be positive). I'll leave verifying that x = y = g produces a minimum as opposed to, say, a maximum or saddle point to the reader.

Last edited: #### uranium235

##### Recruit
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The equality sign should have been a nonstrict inequality. Anyhoo, this reminds me of midpoint (or Jensen) convexity, albeit with the geometric mean instead of the arithmetic one.

Let g = sqrt(ab). We want to prove that p(g)p(g) ≤ p(a)p(b). Let p(x) = c_0 + c_1*x + ... + c_n*x^n. Upon expanding the product p(a)p(b) we get a sum of terms of the form c_i*c_k*a^i*b^k. If we can therefore manage to show that c_i^2*g^(2i) ≤ c_i^2*a^i*b^i & c_i*c_k*2g^(i+k) ≤ c_i*c_k*(a^i*b^k + a^k*b^i) for every i ≠ k, we are done (since a & b & the c_i are all strictly positive). The former inequality is obviously an equality and therefore true. Since the c_i are all strictly positive, the latter inequality is tantamount to showing that 2g^(i+k) ≤ a^i*b^k + a^k*b^i.

Showing that 2g^(i+k) ≤ a^i*b^k + a^k*b^i is easy with Lagrange multipliers. For concreteness, I'll illustrate the case where i = 5 & k = 2. In that case, we want to show that 2g^7 ≤ a^5*b^2 + a^2*b^5 = a^2*b^2*(a^3 + b^3) = g^4*(a^3 + b^3) -- i.e., 2g^3 ≤ a^3 + b^3. Now we use Lagrange multipliers. We want to minimize the bivariate function x^3 + y^3 constrained to the hyperbola xy = g^2. Lagrange multipliers readily yields that 3x^2/y = 3y^2/x -- i.e., x^3 = y^3, so x = y = g (assuming x & y to be positive). I'll leave verifying that x = y = g produces a minimum as opposed to, say, a maximum or saddle point to the reader.
Nice
Here is the latex format for ADHDcels who have difficulty in reading #### uranium235

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Nice
Here is the latex format for ADHDcels who have difficulty in reading

#### Attachments #### Ahnfeltia

##### 3254011055738
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Nice
Here is the latex format for ADHDcels who have difficulty in reading
And here I was wondering what kind of madman would manually transcribe my writings into LaTeX. Nice use of ChatGPT. Replies
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