基本不等式a2+b2≥2ab在不等式的证明中起重要作用,但有些不等式直接用它去证明比较困难,而应用该不等式的变形去证明却比较方便. 变形1a2+b2≥2ab a2+b2≥1/2(a+b)2. 例 1 已知 a,b,c∈R+,且a+b+c=5,a2+b2+c2=9,试证明:1≤a、b、c≤7/3. 证明:由已知 a+b=5-c,a2+b2≥9-c2,∵a2+b2≥1/2(a+b)2,∴9-c2≥1/2(5-c)2,∴3c2-10c+7≤0,∴1≤c≤7/3,同理1≤a≤7/3,1≤b≤7/3. 例2 设a,b∈R+,且a+b=1,求证:(a+1/2)2+(b+1/b)2≥25/2. The basic inequality a2+b2≥2ab plays an important role in the proof of inequality, but some inequality directly use it to prove it is more difficult to prove, but it is more convenient to use the deformation of the inequality to prove. Deformation 1a2+b2≥2ab a2+b2≥1 /2(a+b)2. Example 1 is known as a,b,c∈R+, and a+b+c=5,a2+b2+c2=9. Test proves: 1≤a,b,c≤7 /3. Proof: From the known a+b=5-c,a2+b2≥9-c2,∵a2+b2≥1/2(a+b)2,∴9-c2≥1/2(5- c) 2, ∴ 3c2-10c + 7 ≤ 0, ∴ 1 ≤ c ≤ 7/3, empathy 1 ≤ a ≤ 7/3, 1 ≤ b ≤ 7/3. Example 2 Let a, b ∈ R+, and a+b=1, verification: (a+1/2)2+(b+1/b)2≥25/2.