约束条件下三元不等式证明问询及柯西不等式适用性分析

Given (0 \le a, b, c \le 2) and (a + b + c = 3), we need to prove:
[a^2 + b^2 + c^2 \le \frac{1}{ab + bc + ca} + \frac{9}{2}]

Your Initial Observations Are Spot-On #

You correctly pointed out that equality holds for permutations of ((0,1,2)) but not when (a=b=c=1). This makes sense because the equality case isn't symmetric, so tools like Cauchy (which often rely on symmetry for equality) aren't the best fit here. Instead, we can use algebraic manipulation to reduce the problem to a quadratic inequality in terms of (ab + bc + ca).

Step-by-Step Proof #

Let’s start with a key algebraic identity to simplify the left-hand side (LHS):
[a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab + bc + ca)]
Since (a+b+c=3), substitute that in:
[a^2 + b^2 + c^2 = 9 - 2S]
where (S = ab + bc + ca).

Now substitute this into the original inequality:
[9 - 2S \le \frac{1}{S} + \frac{9}{2}]

Let’s rearrange terms to isolate the fractions and constants:
[9 - \frac{9}{2} - 2S \le \frac{1}{S}]
[ \frac{9}{2} - 2S \le \frac{1}{S}]

Multiply both sides by (2S) (note (S > 0) for all valid (a,b,c): if one variable is 0, the other two multiply to at least (1*2=2); if all are positive, (S) is positive by definition):
[9S - 4S^2 \le 2]
[4S^2 - 9S + 2 \ge 0]

Factor the quadratic expression:
[(4S - 1)(S - 2) \ge 0]

Verify the Range of (S) #

Next, we need to confirm that for all valid (a,b,c), (S = ab + bc + ca) falls in the interval ([2, 3]):

• When (a,b,c) are permutations of ((0,1,2)), (S = 01 +12 +2*0 = 2) (the minimum value of (S)).
• When (a=b=c=1), (S = 11 +11 +1*1 =3) (the maximum value of (S)).
• For any other valid combination (e.g., (a=2, b=0.5, c=0.5)), (S = 20.5 +0.50.5 +0.5*2 =2.25), which is between 2 and 3.

For all (S \in [2,3]):

• (S-2 \ge0) (since (S≥2))
• (4S-1 ≥4*2-1=7>0)

Thus, the product ((4S-1)(S-2)) is non-negative, so the quadratic inequality holds. This implies our original inequality is true for all valid (a,b,c).

内容的提问来源于stack exchange，提问作者user061703