Knowing that a, b, c, and d are all positive numbers, verify ab cd ac bd 4abcd

If a, and b, c, and d are all positive, verify (b a+d c) (c b+a call search d) 4

b a+d c)(c b+a missing d)=c a+d b+b d+a c 4[(c a)(d b)(b d)(a c)] 1 4) = 4 1 (1 4) = 4

The equal sign holds if and only if c a=d b=b d=a c, i.e., a=b=c=d=1.

Divide the three items on the left into two halves, a total of six items, group, and then use the average to not be stupid and equal to the divination.

bc/a+ac/b+ab/c

bc/2a+ac/2b)+(ac/2b+ab/2c)+(bc/2a+ab/2c)

2√(bc/2a×ac/2b)+2√(ac/2b×ab/2c)+2√(bc/2a×ab/2c)

2 (C 4)+2 (A 4)+2 Bring (B 4) C + A + B to complete.

Hope. Thank you.

a,b,c∈r+

By the basic wanton dislike split-hand inequality x 2 + y 2 2xybc 2a) + (ac 2b) 2 [(bc 2a)(ac 2b)]=2 (abc 2 4ab)=c

BC 2a) + (ab and beat 2c) 2 [(BC 2a)(ab 2c)]=2 (ACB 2 4ac)=b

ac/2b)+(ab/2c)≥2√[(ac/2b)(ab/2c)]=2√(bca^2/4bc)=a

Add the three formulas to get it:

bc/a)+(ac/b)+(ab/c)≥a+b+c

Certificate: BC A+AC B+AB C

abc a + abc b + abc finch and c abc (1 a +1 b +1 c).

1/a-1/b)²≥0

1/a²+1/b²≥2/ab

1/b-1/c)²≥0

1/b²+1/c²≥2/bc

1/a-1/b)²≥0

1 A +1 Yuanxun C 2 AC

2 A +2 B +2 C 2 Ab+2 BC+2 BC+2 CA1 A +1 B +1 C 1 AB+1 BC+1 CABC A+AC B+AB C ABC(1 AB+1 BC+1 CA)=A+B+C

bc/a+ac/b+ab/c≥a+b+c

When a, b, and c are positive real numbers that are not equal to each other, bc a+ac b+ab c>a+b+c

If you want to prove that BC A+AC B+AB C>A+B+C, you also need to add the condition that "A, B, and C are not equal to each other".