TargetIIT

If xy+yz+zx = 3, prove that

[im]http://codecogs.izyba.com/gif.latex?%5Csqrt%7B1+x%5E4%7D%20+%20%5Csqrt%20%7B1+y%5E4%7D%20+%20%5Csqrt%20%7B1+z%5E4%7D%20%5Cge%203%20%5Csqrt%202[/im]

[im]http://codecogs.izyba.com/gif.latex?x%3D%5Csqrt%7B3%7Dtan%28%5Cfrac%7BA%7D%7B2%7D%29%20%5C%5C%20y%3D%5Csqrt%7B3%7Dtan%28%5Cfrac%7BB%7D%7B2%7D%29%20%5C%5C%20z%3D%5Csqrt%7B3%7Dtan%28%5Cfrac%7BC%7D%7B2%7D%29%20%5C%5C%20%5Csum%7Bx%5Csqrt%7Bx%5E2%20+%20%5Cfrac%7B1%7D%7Bx%5E2%7D%7D%7D%5Cgeq%20%5Csum%7Bx%7D%5Csqrt%7B2%7D%5Cgeq%20%5Csqrt%7B3%7D%5Csqrt%7B2%7D%5Csum%7Btan%5Cfrac%7BA%7D%7B2%7D%7D%5Cgeq%203%5Csqrt%7B2%7D[/im]

didnt say they are all +ve

x[p]4[/p] , y[p]4[/p] , z[p]4[/p] , are always +ve , independant of x , y , z .

So we can apply am-gm on 1 + x[p]4[/p] , etc.

We get [frac]1 + x[p]4[/p][/]2[/frac] ≥ x[p]2[/p]

or [sqrt]1 + x[p]4[/p]   [/sqrt] ≥ [sqrt]2[/sqrt] x

similarly [sqrt]1 + y[p]4[/p]  [/sqrt] ≥ [sqrt]2[/sqrt] y

and [sqrt]1 + z[p]4[/p]    [/sqrt] ≥ [sqrt]2[/sqrt] z

adding all of them together , we get L.H.S ≥ [sqrt]2[/sqrt] ( x + y + z) ---------------- 1

now ( x+ y + z )[p]2[/p] = x[p]2[/p] + y[p]2[/p] + z[p]2[/p] + ( xy + zx + yz ) ≥ 9 ,--------------A

as given xy + zx + yz = 3 and accordingly each of x[p]2[/p] , y[p]2[/p] , z[p]2[/p] ≥ 1 ( if none of them are zero )

however , even if one or two of x , y , z are zero , then also we get either two or one of x , y , z ≥ 2

so from A , we get x + y + z ≥ 3 ------------------ 2

again from 1 and 2 , L.H.S ≥ [sqrt]2[/sqrt] ( 3 ) ≥ 3[sqrt]2[/sqrt]

As started by Soumya, we arrive at [im]http://codecogs.izyba.com/gif.latex?%5Csqrt%7B1+x%5E4%7D%5Cge%20%5Csqrt%7B2%7Dx[/im], [im]http://codecogs.izyba.com/gif.latex?%5Csqrt%7B1+y%5E4%7D%5Cge%20%5Csqrt%7B2%7Dy[/im], [im]http://codecogs.izyba.com/gif.latex?%5Csqrt%7B1+z%5E4%7D%5Cge%20%5Csqrt%7B2%7Dz[/im]

[im]http://codecogs.izyba.com/gif.latex?%28x+y+z%29%5E2%3Dx%5E2+y%5E2+z%5E2+6[/im]

From Cauchy-Schwarz on [im]http://codecogs.izyba.com/gif.latex?%28a_1%2Ca_2%2Ca_3%29%3D%28x%2Cy%2Cz%29[/im] & [im]http://codecogs.izyba.com/gif.latex?%28b_1%2Cb_2%2Cb_3%29%3D%28y%2Cz%2Cx%29[/im], we arrive at [im]http://codecogs.izyba.com/gif.latex?%28x%5E2+y%5E2+z%5E2%29%5E2%5Cge%209%20%5CRightarrow%20%28x+y+z%29%5E2%5Cge%209[/im]

Now if x+y+z ≥ 3, it's as soumya has done....

But if [im]http://codecogs.izyba.com/gif.latex?%28x+y+z%29%5Cle%20-3[/im], then we write the entire thing - but this time taking the negative square roots like
[im]http://codecogs.izyba.com/gif.latex?%5Csqrt%7B1+x%5E4%7D%5Cge%20-%5Csqrt%7B2%7Dx[/im]

Adding the 3 expressions we have our result.

So its better you write it as [im]http://codecogs.izyba.com/gif.latex?%5Csqrt%7B1+x%5E4%7D%20%5Cge%20%5Csqrt%202%20%7Cx%7C[/im]

I did using CS

[im]http://codecogs.izyba.com/gif.latex?2%281+x%5E4%29%20%5Cge%20%281+x%5E2%29%5E2%20%5CRightarrow%20%5Csqrt%7B2%281+x%5E4%29%7D%20%5Cge%201+x%5E2[/im]

and we have [im]http://codecogs.izyba.com/gif.latex?%5Csum%201+x%5E2%20%3D%203%20+%20%5Csum%20x%5E2%20%5Cge%203%20+%20%5Csum%20xy%20%3D%206[/im]