Metamath Proof Explorer


Theorem young2d

Description: Young's inequality for n = 2 , a direct application of amgmw2d . (Contributed by Kunhao Zheng, 6-Jul-2021)

Ref Expression
Hypotheses young2d.0 ⊢ φ → A ∈ ℝ +
young2d.1 ⊢ φ → P ∈ ℝ +
young2d.2 ⊢ φ → B ∈ ℝ +
young2d.3 ⊢ φ → Q ∈ ℝ +
young2d.4 ⊢ φ → 1 P + 1 Q = 1
Assertion young2d ⊢ φ → A ⁢ B ≤ A P P + B Q Q

Proof

Step Hyp Ref Expression
1 young2d.0 ⊢ φ → A ∈ ℝ +
2 young2d.1 ⊢ φ → P ∈ ℝ +
3 young2d.2 ⊢ φ → B ∈ ℝ +
4 young2d.3 ⊢ φ → Q ∈ ℝ +
5 young2d.4 ⊢ φ → 1 P + 1 Q = 1
6 2 rpred ⊢ φ → P ∈ ℝ
7 1 6 rpcxpcld ⊢ φ → A P ∈ ℝ +
8 2 rpreccld ⊢ φ → 1 P ∈ ℝ +
9 4 rpred ⊢ φ → Q ∈ ℝ
10 3 9 rpcxpcld ⊢ φ → B Q ∈ ℝ +
11 4 rpreccld ⊢ φ → 1 Q ∈ ℝ +
12 7 8 10 11 5 amgmw2d ⊢ φ → A P 1 P ⁢ B Q 1 Q ≤ A P ⁢ 1 P + B Q ⁢ 1 Q
13 2 rpcnd ⊢ φ → P ∈ ℂ
14 2 rpne0d ⊢ φ → P ≠ 0
15 13 14 recidd ⊢ φ → P ⁢ 1 P = 1
16 15 oveq2d ⊢ φ → A P ⁢ 1 P = A 1
17 13 14 reccld ⊢ φ → 1 P ∈ ℂ
18 1 6 17 cxpmuld ⊢ φ → A P ⁢ 1 P = A P 1 P
19 1 rpcnd ⊢ φ → A ∈ ℂ
20 19 cxp1d ⊢ φ → A 1 = A
21 16 18 20 3eqtr3d ⊢ φ → A P 1 P = A
22 4 rpcnd ⊢ φ → Q ∈ ℂ
23 4 rpne0d ⊢ φ → Q ≠ 0
24 22 23 recidd ⊢ φ → Q ⁢ 1 Q = 1
25 24 oveq2d ⊢ φ → B Q ⁢ 1 Q = B 1
26 22 23 reccld ⊢ φ → 1 Q ∈ ℂ
27 3 9 26 cxpmuld ⊢ φ → B Q ⁢ 1 Q = B Q 1 Q
28 3 rpcnd ⊢ φ → B ∈ ℂ
29 28 cxp1d ⊢ φ → B 1 = B
30 25 27 29 3eqtr3d ⊢ φ → B Q 1 Q = B
31 21 30 oveq12d ⊢ φ → A P 1 P ⁢ B Q 1 Q = A ⁢ B
32 7 rpcnd ⊢ φ → A P ∈ ℂ
33 32 13 14 divrecd ⊢ φ → A P P = A P ⁢ 1 P
34 10 rpcnd ⊢ φ → B Q ∈ ℂ
35 34 22 23 divrecd ⊢ φ → B Q Q = B Q ⁢ 1 Q
36 33 35 oveq12d ⊢ φ → A P P + B Q Q = A P ⁢ 1 P + B Q ⁢ 1 Q
37 36 eqcomd ⊢ φ → A P ⁢ 1 P + B Q ⁢ 1 Q = A P P + B Q Q
38 12 31 37 3brtr3d ⊢ φ → A ⁢ B ≤ A P P + B Q Q