Toán Lớp 9: CMR: căn(4-a^2)+ căn(4-b^2)+ căn(4-c^2) ≤ 3căn3 với a,b,c ∈(2;-2) thoả mãn a+b+c=3

Question

Toán Lớp 9:  CMR: căn(4-a^2)+ căn(4-b^2)+ căn(4-c^2) ≤ 3căn3  với a,b,c ∈(2;-2) thoả mãn a+b+c=3

dananhthumai · Answer

$ begin{array}{l} A =  sqrt {4 - {a^2}}  +  sqrt {4 - {b^2}}  +  sqrt {4 - {c^2}}    A =  sqrt { left( {2 - a}  right) left( {a + 2}  right)}  +  sqrt { left( {2 - b}  right) left( {2 + b}  right)}  +  sqrt { left( {2 - c}  right) left( {c + 2}  right)}      Rightarrow  sqrt 3 A =  sqrt {3 left( {2 - a}  right) left( {a + 2}  right)}  +  sqrt {3 left( {2 - b}  right) left( {2 + b}  right)}  +  sqrt {3 left( {2 - c}  right) left( {c + 2}  right)}      Rightarrow  sqrt 3 A =  sqrt { left( {6 - 3a}  right) left( {a + 2}  right)}  +  sqrt { left( {6 - 3b}  right) left( {2 + b}  right)}  +  sqrt { left( {6 - 3c}  right) left( {c + 2}  right)}    a,b,c  in  left( { - 2;2}  right)  Rightarrow  left {  begin{array}{l} a + 2 > 0   6 - 3a > 0  end{array}  right.   AM - GM  Leftrightarrow  left( {6 - 3a}  right) left( {2 + a}  right)  le  dfrac{{{{ left( {6 - 3a + a + 2}  right)}^2}}}{4}     Leftrightarrow  sqrt { left( {6 - 3a}  right) left( {2 + a}  right)}   le  dfrac{{8 - 2a}}{4} = 4 - a   Tương , tự: sqrt { left( {6 - 3b}  right) left( {2 + b}  right)}   le 4 - b, sqrt { left( {6 - 3c}  right) left( {4 + c}  right)}   le 4 - c     Rightarrow  sqrt 3 A  le 4 + 4 + 4 -  left( {a + b + c}  right) = 12 - 3 = 9     Rightarrow A  le 3 sqrt 3    ' = '  Leftrightarrow a = b = c = 1  end{array}$