Metamath Proof Explorer


Theorem abstri

Description: Triangle inequality for absolute value. Proposition 10-3.7(h) of Gleason p. 133. (Contributed by NM, 7-Mar-2005) (Proof shortened by Mario Carneiro, 29-May-2016)

Ref Expression
Assertion abstri ABA+BA+B

Proof

Step Hyp Ref Expression
1 2re 2
2 1 a1i AB2
3 simpl ABA
4 simpr ABB
5 4 cjcld ABB
6 3 5 mulcld ABAB
7 6 recld ABAB
8 2 7 remulcld AB2AB
9 abscl AA
10 3 9 syl ABA
11 abscl BB
12 4 11 syl ABB
13 10 12 remulcld ABAB
14 2 13 remulcld AB2AB
15 10 resqcld ABA2
16 12 resqcld ABB2
17 15 16 readdcld ABA2+B2
18 releabs ABABAB
19 6 18 syl ABABAB
20 absmul ABAB=AB
21 3 5 20 syl2anc ABAB=AB
22 abscj BB=B
23 4 22 syl ABB=B
24 23 oveq2d ABAB=AB
25 21 24 eqtrd ABAB=AB
26 19 25 breqtrd ABABAB
27 2rp 2+
28 27 a1i AB2+
29 7 13 28 lemul2d ABABAB2AB2AB
30 26 29 mpbid AB2AB2AB
31 8 14 17 30 leadd2dd ABA2+B2+2ABA2+B2+2AB
32 sqabsadd ABA+B2=A2+B2+2AB
33 10 recnd ABA
34 12 recnd ABB
35 binom2 ABA+B2=A2+2AB+B2
36 33 34 35 syl2anc ABA+B2=A2+2AB+B2
37 15 recnd ABA2
38 14 recnd AB2AB
39 16 recnd ABB2
40 37 38 39 add32d ABA2+2AB+B2=A2+B2+2AB
41 36 40 eqtrd ABA+B2=A2+B2+2AB
42 31 32 41 3brtr4d ABA+B2A+B2
43 addcl ABA+B
44 abscl A+BA+B
45 43 44 syl ABA+B
46 10 12 readdcld ABA+B
47 absge0 A+B0A+B
48 43 47 syl AB0A+B
49 absge0 A0A
50 3 49 syl AB0A
51 absge0 B0B
52 4 51 syl AB0B
53 10 12 50 52 addge0d AB0A+B
54 45 46 48 53 le2sqd ABA+BA+BA+B2A+B2
55 42 54 mpbird ABA+BA+B