21xy+yz+zx \le x^2+y^2+z^2
\Rightarrow xy+yz+zx \le 3
If x,y,z≥0 and x2+y2+z2=3
Find the max for the expression
F=xy+xz+yz+5x+y+z
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5 Answers
62given expression
=\\\frac{(x+y+z)^2-x^2-y^2-z^2}{2}+\frac{5}{x+y+z} \\=(x+y+z)^2/2 + 5/(x+y+z)-3/2
Now x+y+z=t and t>0
we have to maximize t^2/2+5/t-3/2
which u can do by AM GM or by maxima minima...
for AM GM you will write t^2/2+5/2t+5/2t-3/2
applying AM GM on the first three terms, the given expression >= 3*3√25/8-3/2
1Pardon, The required is the minimum, However
The answ from Nishant is correct.
21f(t) = t^2/2+5/t-3/2
f'(t) = t-5/t2
f'(t)≤0 when t≤(5)(1/3)
f'(t)≥0 when t≥(5)(1/3)
my means inequality ((x2+ y2+ z2)/3)(1/2)≥(x+y+z)/3
giving 3≥(x+y+z)
(x+y+z)2= 3+ 2(xy+yz+x)≥3 (with equality iff two terms r zero)
(x+y+z)≥ √3
f is decreasing in the interval (0,(5)(1/3)) and increasing ((5)(1/3),∞)
note that √3<(5)(1/3)
Hence maxf(x+y+z) = max{f(3),f(√3)} = f(3) = 3+ 5/3
21To be precise f(x+y+x) is convex , hence for max we need to check only in the end points of its domain.
ps. i m not well now may be i m wrong..pls tell me....its ri8 or wrong
just now i wrote dx/x+1 = arctanx :D