`a+b+c=6` `0 ≤a;b;c ≤4` tìm `maxP=a^2+b^2+c^2+ab+ac+bc`

Ta có:

P=a2+b2+c2+ab+ac+bcP=a2+b2+c2+ab+ac+bc

=(a+b+c)2−2ab−2bc−2ac+ab+ac+bc=(a+b+c)2-2ab-2bc-2ac+ab+ac+bc

=62−ab−ac−bc=62-ab-ac-bc

=36−(ab+ac+bc)=36-(ab+ac+bc)

Vì: a;b;c≤4a;b;c≤4

⇒(a−4)(b−4)(c−4)≤0⇒(a-4)(b-4)(c-4)≤0

⇔abc−4ac−4bc−64−4ab+16a+16b+16c≤0⇔abc-4ac-4bc-64-4ab+16a+16b+16c≤0

⇔abc−4(ab+ac+bc)+16(a+b+c)−64≤0⇔abc-4(ab+ac+bc)+16(a+b+c)-64≤0 

Vì: 0≤a;b;c0≤a;b;c

⇒abc≥0⇒abc≥0

⇔−abc≤0⇔-abc≤0 

Cộng 22 BĐT cùng chiều theo từng vế, ta được:

abc−4(ab+ac+bc)+16(a+b+c)−64−abc≤0abc-4(ab+ac+bc)+16(a+b+c)-64-abc≤0 

⇔−4(ab+ac+bc)+16(a+b+c)≤64⇔-4(ab+ac+bc)+16(a+b+c)≤64

⇔−4(ab+ac+bc)+16.6≤64⇔-4(ab+ac+bc)+16.6≤64

⇔−4(ab+ac+bc)≤−32⇔-4(ab+ac+bc)≤-32

⇔−(ab+ac+bc)≤−8⇔-(ab+ac+bc)≤-8

⇒36−(ab+ac+bc)≤36−8=28⇒36-(ab+ac+bc)≤36-8=28

Vậy maxpmaxp là 2828 khi: (a;b;c)=(0;2;4)(a;b;c)=(0;2;4) và các hoán vị