It is known that a, B and C are the side lengths of the triangle ABC, and B ^ 2 + 2Ab = C ^ 2 + 2Ac

It is known that a, B and C are the side lengths of the triangle ABC, and B ^ 2 + 2Ab = C ^ 2 + 2Ac

b^2+2ab=c^2+2ac
b2-c2=2a(c-b)
(b-c)(b+c)=-2a(b-c),
Suppose b > = C
Left (B-C) (B + C) > = 0, right-2a (B-C) 0, a > 0,
Only B-C = 0 is possible
B-C = 0, B = C, isosceles
That is, the original triangle is isosceles triangle

It is known that a, B, C are the three side lengths of △ ABC. When B2 + 2Ab = C2 + 2Ac, try to judge which kind of triangle △ ABC belongs to and explain the reason

∵b2+2ab=c2+2ac，
∴b2+2ab+a2=c2+2ac+a2，
∴（b+a）2=（c+a）2，
∵ a, B, C are the three sides of ᙽ ABC,
/ / A, B and C are all positive numbers,
∴b+a=c+a，
∴b=c，
The triangle is an isosceles triangle

In this paper, we prove that the theorem of "a C + a" is true

Extend ad to e so that de = ad, connect be
∵ ad is the center line on the side of BC
∴∠ADC=∠BDE,BD=CD
∴⊿ADC≌⊿BDE(SAS)
∴AC=BE
∵AD+BD＞AB
DE+BD＞BE
∴AD+DE+2BD＞AB+BE
∴2AD+2BD＞AB+BE=AB+AC
AD + BD > 1 / 2 (AB + AC)

It is known that in the triangle ABC, ad is the center line on the edge of BC. This paper tries to explain the reason why the inequality AD + BD is greater than 1 / 2 (AB + AC)

AD + BD > AB (in triangle ADB)
AD + DC > AC (in triangle ACD)
Because BD = DC
The two sides of the inequality are added separately
2(AD+BD)>AB+AC
Divide both sides by 2 to get AD + BD > 1 / 2 (AB + AC)

It is known that in △ ABC, ad is the midline on the edge of BC 2（AB+AC）．

It is proved that: BD + ad > AB, CD + ad > AC,
∴BD+AD+CD+AD＞AB+AC．
∵ ad is the center line on the edge of BC, BD = CD,
∴AD+BD＞1
2（AB+AC）．

Proof of inequality: a + 1 / a - √ (a 2 + 1 / a 2) ≤ 2 - √ (2)

It's too much trouble upstairs. Look at me
Let a + 1 / a = a
y=(a+1/a)-√(a^2+1/a^2)=(a+1/a)-√[(a+1/a)^2-2]=A-√(A^2-2)=2/[A+√(A^2-2)] A>=2.
Obviously, the maximum value of Y on a > = 2 is obtained when a = 2, y = 2 / (2 + √ 2) = 2 - √ 2

Univariate quadratic inequality x? - x + 1 > 1 / 3x (x-1)

x²-x-1＞1/3（x²-x）
Let t = x? - x, then t + 1 > 1 / 3T
In this paper, we discuss the case that t is greater than 0 and less than 0, simplify the inequality, find out the range of T and then substitute it into the range of X

What is the solution set of inequality x 2 - | x | 0

[reference answer]
① When x ≥ 0, the original inequality is
x²-x>0
x(x-1)>0
x> 1 or x1
② When X0
x(x+1)>0
X0
ν X1 or X

Inequality proof of AB > A + B-1 under a ^ 2 + B ^ 2 + 1 / radical sign

To prove the above formula, it is proved that a ^ 2 + B ^ 2 + 1 > AB * (a + b) - AB under the root sign is obtained according to the inequality string. AB * (a + b) - AB under the root sign is greater than or equal to 2Ab - because a ^ 2 + B ^ 2 is greater than or equal to 2Ab, a ^ 2 + B ^ 2 + 1 is greater than AB * (a + b) - AB under root sign

Prove inequality: 2 / (1 / A + 1 / b) ≤ radical ab ≤ (a + b) / 2 ≤ radical ((a ^ 2 + B ^ 2) / 2) (a, B belong to positive real numbers)

in my submission:
So a 2 + B 2 - 2 ab = (a-b) 2 ≥ 0, so a 2 + B 2 ≥ 2 ab is (a 2 + B 2) / 2 ≥ ab
Because a and B belong to positive real numbers, the root sign ((a 2 + B 2) / 2) ≥ the root sign ab
ab - 4/（1/a+1/b）² = （a/b+b/a- 2）/（1/a+1/b）² =(√a/√b-√b/√a)² / （1/a+1/b）² ≥0
Because a and B belong to positive real numbers, 2 / (1 / A + 1 / b) ≤ ab
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