a,b,c>=0
a^2+b^2+c^2=3
=>a,c,b ko đồng thời bằng 0
\(\left|a+b+c\right|< =\sqrt{\left(a^2+b^2+c^2\right)\left(1+1+1\right)}\)
=>-3<=a+b+c<=3
=>0<=a+b+c<=3
P=a+b+c
=>P^2=a^2+b^2+c^2+2(ab+ac+bc)
=3+2(ab+bc+ac)>=3
=>P>căn 3
=>a,b,c thuộc (0;2)
Đặt a=2*cosA, b=2*cosB, c=2*cosC
=>4*cos^2A+4*cos^2B+4*cos^2C=3
=>cos^2A+cos^2B+cos^2C=3/4
=>3/4+2*cosA*cosB*cosC=1
=>2*cosA*cosB*cosC=1/4
=>cosA*cosB*cosC=1/8
\(a+b+c=4\cdot cos\left(\dfrac{A+B}{2}\right)\cdot cos\left(\dfrac{A-B}{2}\right)+2\left(1-2\cdot sin^2\left(\dfrac{C}{2}\right)\right)\)
\(=4\cdot cos\left(\dfrac{A+B}{2}\right)\cdot cos\left(\dfrac{A-B}{2}\right)-4\cdot sin^2\left(\dfrac{C}{2}\right)+2\)
a+b+c<=3
=>\(-\left(2sin\left(\dfrac{C}{2}\right)-cos\left(\dfrac{A-B}{2}\right)\right)^2-sin^2\left(\dfrac{A-B}{2}\right)< =0\)(luôn đúng)
Dấu = xảy ra khi A=B=C=60 độ
=>a=b=c=1
\(P_{max}=\dfrac{1}{4-1\cdot1}+\dfrac{1}{4-1\cdot1}+\dfrac{1}{4-1\cdot1}=1\)
a+b+c>=0
Giả sử a=căn 3, b=c=0
=>\(P_{min}=\dfrac{\sqrt{3}}{4-0}+\dfrac{0}{4-0}+\dfrac{0}{4-0}=\dfrac{\sqrt{3}}{4}\)
