In triangle ABC, the inequality 1 / A + 1 / B + 1 / C > = 9 / π holds; in quadrilateral ABCD, inequality 1 / A + 1 / B + 1 / C + 1 / d > = 8 / π holds The whole problem is: in the triangle ABC, the inequality 1 / A + 1 / B + 1 / C > = 9 / π holds; in the quadrilateral ABCD, the inequality 1 / A + 1 / B + 1 / C + 1 / d > = 8 / π holds; in the Pentagon ABCDEF, the inequality 1 / A + 1 / B + 1 / C + 1 / D + 1 / E > = 25 / 3 π holds; in the n-polygon a1a2a3 Is there an inequality in an? The question is, "what is the inequality?"

In the n-polygon, there are those which add up to ≥ N 2 / [(n-2) π]

In the triangle ABC, the inequality 1 / A + 1 / B + 1 / C > = 9 / π holds; in the triangle ABCD, the inequality 1 / A + 1 / B + 1 / C + 1 / d > = 8 / π holds

In the n-polygon, there are those which add up to ≥ N 2 / [(n-2) π]

In the triangle ABC, a (3, - 1), B (- 3,1), C (1,3), write the system of bivariate linear inequalities represented by the triangle ABC region Including the boundary, the process

0

Here's a premise: A, B, C are all greater than - 1
Firstly, the inequality 1 / (a + 1) + 1 / (B + 1) + 1 / (c + 1) ≥ 3 / {[(a + 1) (B + 1) (c + 1)] ^ (1 / 3)}
While (a + 1) (B + 1) (c + 1) ≤ [(a + 1) + (B + 1) + (c + 1)] 3 / 27 ≤ 6 / 27 = 8
And this is the denominator, so 3 / {[(a + 1) (B + 1) (c + 1)] ^ (1 / 3)} ≥ 3 / [8 ^ (1 / 3)] = 3 / 2

An inequality problem in high school Proof 3 (1 + A ^ 2 + A ^ 4) > = (1 + A + A ^ 2) ^ 2

1 + A ^ 2 + A ^ 4 can be factorized into (1 + A + A ^ 2) (1-A + A ^ 2)
1 + A + A ^ 2 = 3 / 4 + (1 / 2 + a) ^ 2 > 0, both sides can be eliminated at the same time
It is only necessary to prove that 3 (1-A + A ^ 2) ≥ 1 + A + A ^ 2
This is simple, left right =2a^2-4a+2=2 (A-1) ^2 ≥ 0

Let a, B, C be nonzero real numbers, and a + B + C ≠ 0, if a + B − C c＝a−b+c b＝−a+b+c a. Then (a + b) (B + C) (c + a) ABC is equal to () A. 8 B. 4 C. 2 D. 1

∵a+b−c
c＝a−b+c
b＝−a+b+c
a，
∴a+b−c+a−b+c−a+b+c
a+b+c=1=a+b−c
c＝a−b+c
b＝−a+b+c
a，
∴2a=b+c，2c=a+b，2b=a+c，
∴(a+b)(b+c)(c+a)
abc=2c×2a×2b
abc=8，
Therefore, a

Let ABC be a real number, and a ratio B equal to B ratio C equal to C ratio a, find a + B-C ratio A-B + C

From a / b = B / C = C / a =,
B ^ 2 = AC, C ^ 2 = ab
If you divide the two formulas, you get,
b^2/c^2=c/b,
B ^ 3 = C ^ 3,
So B = C,
A / b = B / b = 1,
So a = B,
So a = b = C,
So + C-B ratio
=(a+a-a)/(a-a+a)
=1

Triangle ABC, known a ^ 3 + B ^ 3 = C ^ 3, prove that C > π / 3 This website is another similar question that you explain for netizens. You think that C is not enough to compare with π / 3 I have a method of proof. What's wrong a^3+b^3=(a+b)(a^2+b^2-ab)=c^3 Again, the triangle a + b > C holds a^2+b^2-ab

Your derivation is OK,
At that time, I have already made a comment on it
actually,
∵a^3+b^3=c^3
C is the largest side, then C is the largest angle
∴C＞60º

If the three sides a, B and C of △ ABC satisfy the condition (a-b) (A2 + b2-c2) = 0, then △ ABC is () A. Isosceles triangle B. Right triangle C. Isosceles triangle or right triangle D. Isosceles right triangle

∵（a-b）（a2+b2-c2）=0，
A = B or A2 + B2 = C2
If only a = B holds, it is an isosceles triangle
When only the second condition holds: a right triangle
When two conditions hold simultaneously: isosceles right triangle
Therefore, C

In RT △ ABC, ∠ ACB = 90 °, M is the midpoint of BC, CN ⊥ am is n, and the relationship between

According to the projective theorem (proved by using trigonometric function), cm ^ 2 = Mn * ma, so BM ^ 2 = Mn * ma, so BM / BN = am / BM, so △ BMN ~ △ ABM, so the two angles are equal