Solving the fourth order determinant! A 100 - 1 B 100 - 1 C 100 - 1 D Four in a row~

Solving the fourth order determinant! A 100 - 1 B 100 - 1 C 100 - 1 D Four in a row~

R1 + AR20 1 + AB a 0-1 B 100 - 1 C 100 - 1 D expand 1 + AB a 0-1 C 10 - 1 DC3 + DC21 + AB a ad-1 C1 + cd0 - 10 expand 1 + ab ad-1 1 + CD determinant = (1 + AB) (1 + CD) + ad = 1 + AB + CD + AD + ABCD

ABC is a positive real number, satisfying the condition a + B + C = 1, then the maximum value of a ^ 2 + B ^ 2 + C ^ 2 + 2 (3ABC) ^ (1 / 2) is?

The maximum is 1
a²+b²+c²+2√3·√(abc)
= a²+b²+c²+2√(3abc(a+b+c))
= a²+b²+c²+2√(3((ab)(bc)+(bc)(ca)+(ca)(ab)))
≤ A & # 178; + B & # 178; + C & # 178; + 2 √ ((AB + BC + Ca) & # 178;) (in the inequality (x + y + Z) & # 178; ≥ 3 (XY + YZ + ZX), take x = AB, y = BC, z = CA)
= a²+b²+c²+2(ab+bc+ca)
= (a+b+c)²
= 1.
When a = b = C = 1 / 3, the equal sign holds, so the maximum value is 1

"A ≥ 18" is a necessary and sufficient condition for ∀ positive real number x, 2x + ax ≥ C, then the real number C=______ ．

If C ﹤ 0, then a ≥ 0, not in line with the meaning of the question, if C ﹥ 0, ax ≥ C − 2x, according to X is a positive number, there is a ≥ cx-2x2 ∵ y = cx-2x2, when x is a positive number, the range is y ≤− 2 × & nbsp; (C4) 2 & nbsp; + C × C4 = C28, then a ≥ C28, then C28 = 18 ﹥ C = 1, so the answer is: 1