Question

if x, y, z are real numbers such that x+y+z=6 and xy+yz+zx =3 then what is largest value that x can have

Answer 1

x is the largest value

Answer 2

Using (x+y+z)2=x2+y2+z2+2(xy+yz+zx)(x+y+z)2=x2+y2+z2+2(xy+yz+zx)
So x2+y2+z2=(x+y+z)2−2(xy+yz+zx)=52−2⋅3=19x2+y2+z2=(x+y+z)2−2(xy+yz+zx)=52−2⋅3=19So y2+z2=19−x2y2+z2=19−x2 and y+z=5−xy+z=5−xNow Using CauchySchwarzCauchySchwarz Inequality(y2+z2)(12+12)≥(y+z)2(y2+z2)(12+12)≥(y+z)2So(19−x2)⋅2≥(5−x)2(19−x2)⋅2≥(5−x)2So25+x2−10x≤38−2x225+x2−10x≤38−2x2So3x2−10x−13≤03x2−10x−13≤0So3x2−13x+3x−13≤03x2−13x+3x−13≤0So(x+1)(3x−13)≤0⇒−1≤x≤13/3