3a2+b2+c2=1
2(a2+b2+c2)>=(a+b+c)2.
Therefore 2>=(a+b+c)2.
2>=1+2(ab+bc+ca)
1>=2(ab+bc+ca)
1/2>=(ab+bc+ca)
So, (ab+bc+ca)<=1/2.
Is it right???????????????
1.If a^2+b^2=2 and a,b are real.Find the interval of a+b.
2.If a^2+b^2+c^2=1 then find the interval of ab+bc+ca.
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16 Answers
62a2+b2=2
a2+b2>=2ab
2a2+2b2>=(a+b)2
Hence, (a+b)2<= 4
hence (a+b) lies between -2 and 2
now try the 2nd part yourself.. ?
23a2 +b2 +c2 + 2(ab+bc+ac) = (a+b+c)2
(a+b+c)2 ≥0
so a2 + b2+c2 + 2(ab+bc+ac) ≥ 0
1 ≥-2(ab+bc+ac)
(ab + bc+ ac) ≥ -1/2
62are you sure?
what wen a=1, b=0, c=0
is your condition being satisfied?
362no swastika.. there is a mistake..
how did you reach the conclusion that a2+b2+c2>=2abc ??
3Sorry sir.......
I have done the correction.
now is it right???????????????????
62no swastika.. i am afraid not...
can you give some justification ?
1arshad is right ,i think...m also getting the same answer.........sagnik,has arshad given the right answer??
112{(a-b)2+(b-c)2+(c-a)2} ≥ 0
or, a2+b2+c2-ab-bc-ca ≥ 0
or, ab+bc+ca ≤ 1
(a+b+c)2 ≥ 0
a2+b2+c2+2(ab+bc+ca) ≥ 0
ab+bc+ca ≥ -12
Interval : [-12 , 1]
1708(1).If a2+b2=2 and a,b are real.Find the interval of a+b.
Ans: Here we can take a = √2cos p
and b = √2 sin p
so a+b = √2(cos p+sin p)
and we now that -√2≤ cos p + sin p ≤√2
so Min (a+b) = -2 and
Max (a+b) = 2
23are sry sry....maine last ka inequality sign ulta dal diya ..corrected it
