Algebra 3671
2±x+x^2 ±x^3 +x^4 ±···±x^2 k−^1 +x^2 k> 0 ,for all 2kchoices of the signs+and−. This reduces to
(
1
2±x+1
2
x^2)
+
(
1
2
x^2 ±x^3 +1
2
x^4)
+···+
(
1
2
x^2 k−^2 ±x^2 k−^1 +1
2
x^2 k)
+
1
2
x^2 k> 0 ,which is true because^12 x^2 k−^2 ±x^2 k−^1 +^12 x^2 k=^12 (xk−^1 ±xk)^2 ≥0 and^12 x^2 k≥0, and
the equality cases cannot hold simultaneously.
103.This is the Cauchy–Schwarz inequality applied to the numbersa 1 =a√
b, a 2 =
b√
c, a 3 =c√
aandb 1 =c√
b, b 2 =a√
c, b 3 =b√
a. Indeed,9 a^2 b^2 c^2 =(abc+abc+abc)^2 =(a 1 b 1 +a 2 b 2 +a 3 b 3 )^2
≤(a 12 +a^22 +a 32 )(b^21 +b^22 +b 32 )=(a^2 b+b^2 c+c^2 a)(c^2 b+a^2 c+b^2 a).104.By the Cauchy–Schwarz inequality,(a 1 +a 2 +···+an)^2 ≤( 1 + 1 +···+ 1 )(a^21 +a 22 +···+a^2 n).Hencea^21 +a 22 +···+an^2 ≥n. Repeating, we obtain(a 12 +a 22 +···+a^2 n)^2 ≤( 1 + 1 +···+ 1 )(a^41 +a 24 +···+an^4 ),which shows thata^41 +a 24 +···+an^4 ≥n, as desired.
105.Apply Cauchy–Schwarz:(a 1 aσ(a)+a 2 aσ( 2 )+···+anaσ (n))^2 ≤(a^21 +a 22 +···+a^2 n)(aσ( 1 )+aσ( 2 )+···+a^2 σ (n))
=(a^21 +a 22 +···+an^2 )^2.The maximum isa 12 +a^22 +···+an^2. The only permutation realizing it is the identity
permutation.
106.Applying the Cauchy–Schwarz inequality to the numbers√
f 1 x 1 ,√
√ f^2 x^2 ,...,
fnxnand√
f 1 ,√
f 2 ,...,√
fn, we obtain(f 1 x^21 +f 2 x^22 +···+fnxn^2 )(f 1 +f 2 +···+fn)≥(f 1 x 1 +f 2 x 2 +···+fnxn)^2 ,hence the inequality from the statement.