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

It is known that a, B, C are the lengths of the three sides of △ ABC. When B ^ 2 = 2Ab = C ^ 2 + 2Ac, what kind of triangle does △ ABC belong to

This is the isosceles triangle a = B

The proof of a mathematical inequality, known - C / a ＜ - D / B, BC > ad

-C / a ＜ - D / B is
c/a>d/b,
Multiply both sides by ab to get:
bc>ad,
The unequal sign does not change direction
It shows that ab > 0
It can also be proved by the method of proof to the contrary
If ab ≤ 0
Then we know that C / a > D / b
Multiply both sides by ab to get:
bc≤ad
This is contrary to the known BC > ad
So: ab > 0

[senior one mathematics] related inequality proof: known a > b, ab = 1, verification: a 2 + B 2 ≥ 2 √ 2 (a-b) When a ＞ B, ab = 1, it is proved that a  2 + B  ≥ 2 √ 2 (a-b)

(a²+b²)/(a-b)
=[(a-b)²+2ab]/(a-b)
=(a-b)+[2/(a-b)]≥2√2
The minimum value of the original formula = 2 √ 2

Proof of inequality AB = 1 proving a ^ 2 + B ^ 2 > = 2 radical sign 2 (a-b) The product of radical 2 and (a-b)

∵a^2 +b^2 ≥2√2（a - b）
∴（a-b)^2 +2 ≥2√2（a - b）
Let x = A-B, then x ^ 2 - 2 √ 2x + 2 ≥ 0, that is (x - √ 2) ^ 2 ≥ 0
∵ (x - √ 2) ^ 2 ≥ 0 always holds ᙽ the original question is proved

It is proved that (1) if ABC = 1, then (2 + a) (2 + b) (2 + C) is greater than or equal to 27

Because 2 + a = 1 + 1 + a ≥ 3 √ a, and the equal sign holds if and only if a = 1
Similarly, 2 + B ≥ 3 √ B, 2 + C ≥ 3 √ C
So: (2 + a) (2 + b) (2 + C) ≥ 27 3 √ ABC = 27, the equal sign holds if and only if a = b = C = 1

It is proved that there is ABC ^ 3 for any positive number a, B, C

This is a typical application of Lagrange multiplier method
Consider the maximum value problem of F (x, y, z) = x ^ 2Y ^ 2Z ^ 6 under the condition of x ^ 2 + y ^ 2 + Z ^ 2 = 5R ^ 2. Consider only the case where x, y, Z are greater than 0, let a be a multiplier,
Let f (x, y, Z, a) = f (x, y, z) + a (x ^ 2 + y ^ 2 + Z ^ 2 -- 5R ^ 2), and consider the three equations with partial derivatives of 0, it is easy to draw a conclusion that:
X ^ 2 = y ^ 2 = Z ^ 2 / 3, so it is easy to know that the maximum point is reached at x = R, y = R, z = radical (3) r, that is, there is
x^2y^2z^5

Inequality proves that a, B, C are unequal positive numbers, and ABC = 1, prove √ a + √ B + √ C

prove:
Because 1 / A + 1 / b > 2 √ (1 / AB) = 2 √ (ABC / AB) = 2 √ C,
1/a+1/c>2√b
1/b+1/c>2√a
Sum of three forms
SO 2 (1 / A + 1 / B + 1 / C) > 2 (√ a + √ B + √ C)
That is √ a + √ B + √ C

Let a, B, C belong to R +. It is proved that: (a ^ a) * (b ^ b) * (C ^ C) ≥ (ABC) ^ ((a + B + C) / 3)

Take the logarithm to prove 3 (alna + blnb + clnc) > = (a + B + C) (LNA + LNB + LNC), the result is: alna + blnb + clnc > = AlNb + blnc + clnaalna + blnb + clnc > = ALNC + blna + clnbb + clnc = alna + blnb + clnc, then 3 (alna + blnb + clnc) > = (a + B + C) (LNA + LNB +...)

Proof of mathematical inequality a. B, C are the three sides of a triangle, and it is proved that a? / (2B? + 2C? - a?) + B? / (2C? + 2A? - B?) + C? / (2a? + 2B? - C?) > = 1

Let x = 2B? + 2C? - a? > = (B + C) ^ 2-A ^ 2 > 0 (a, B, C are the three sides of a triangle), let y = 2C? + 2A? - B? Z = 2A? + 2B? - C
Then x, y, z > 0, a ^ 2 = (2Y + 2Z-X) / 9 B ^ 2 = (2x + 2z-y) / 9 C ^ 2 = (2x + 2y-z) / 9 is taken into the left formula = 2 / 9 (Y / x + X / y + Z / X + X / Z + Y / Z + Z / y) - 1 / 3 > = 12 / 9-1 / 3 = 1

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