نابرابری با شرط a+b+c=1
[font=Arial,sans-serif]میدانیم [/font] [(a+(b-c)^2/4)+(b+c-(b-c)^2/4)][1+2]>=[radical(a+(b-c)^2/4)+radical(2(b+c)-(b-c)^2/2)]^2](=ـ 3)
radical(2(b+c)-(b-c)^2/2)>=radical(b)+radical(c) z[font=Arial,sans-serif]کافیست ثابت کنیم [/font]
[font=Arial,sans-serif] [/font] z 2(b+c)-(b-c)^2/2>=[radical(b)+radical(c)]^2
[font=Arial,sans-serif] [/font] (2b+2c-b^2/2-c^2/2+bc>=b+c+2*radical(bc
[font=Arial,sans-serif] [/font] B+c-2*radical(bc)>=b^2/2+c^2/2-bc
[font=Arial,sans-serif] [/font] radical(b)-radical(c)]^2>=(b-c)^2/2]
[font=Arial,sans-serif]همیشه برقرار است [/font]b,c<=1[font=Arial,sans-serif] که با توجه به [/font] [font=Arial,sans-serif].[/font]
radical(2(b+c)-(b-c)^2/2)>=radical(b)+radical(c) z[font=Arial,sans-serif]کافیست ثابت کنیم [/font]
[font=Arial,sans-serif] [/font] z 2(b+c)-(b-c)^2/2>=[radical(b)+radical(c)]^2
[font=Arial,sans-serif] [/font] (2b+2c-b^2/2-c^2/2+bc>=b+c+2*radical(bc
[font=Arial,sans-serif] [/font] B+c-2*radical(bc)>=b^2/2+c^2/2-bc
[font=Arial,sans-serif] [/font] radical(b)-radical(c)]^2>=(b-c)^2/2]
[font=Arial,sans-serif]همیشه برقرار است [/font]b,c<=1[font=Arial,sans-serif] که با توجه به [/font] [font=Arial,sans-serif].[/font]
