设a,b,c是△ABC的三边长，求证:
a^2/(a^2+b^2)+b^2/(b^2+c^2)+c^2/(c^2+a^2)>7/5.

设a,b,c是三角形的三边长，求证:
a^2/(a^2+b^2)+b^2/(b^2+c^2)+c^2/(c^2+a^2)>7/5. (1)

(1)式两边同乘以2减3得:
2a^2/(a^2+b^2)+2b^2/(b^2+c^2)+2c^2/(c^2+a^2)-3>14/5-3.
<==>
(a^2-b^2)/(a^2+b^2)+(b^2-c^2)/(b^2+c^2)+(c^2-a^2)/(c^2+a^2)>-1/5
<==>
(a^2-b^)(b^2-c^2)(c^2-a^2)/(a^2+b^)(b^2+c^2)(c^2+a^2)>-1/5
<==>
(a^2+b^)(b^2+c^2)(c^2+a^2)>-5(a^2-b^)(b^2-c^2)(c^2-a^2)

设a>b>c，则只需证
(a^2+b^)(b^2+c^2)(c^2+a^2)>5(a^2-b^)(b^2-c^2)(a^2-c^2) (2)

设x，y，z为正实数，则令:a=y+z，b=z+x，c=x+y，
因为a>b>c，则x<y<z。对(2)作置换得:
[(y+z)^2+(z+x)^2]*[(z+x)^2+(x+y)^2]*[(x+y)^2+(y+z)^2]>
5[(y+z)^2-(z+x)^2]*[(z+x)^2-(x+y)^2]*[(y+z)^2-(x+y)^2]

上式展开化简整理为:
2x[x^5+x^4*(8y-2z)+x^3*(18y^2+12yz-7z^2)+x^2*(6y^3+30y^2*z+10z^2*y+6z^3)+x*(-7y^4+10y^3*z+28y^2*z^2+30z^3*y+18z^4)-2y^5+12y^4*z+30y^3*z^2+10y^2*z^3+12z^4*y+8z^5]+2(z^6-2y*z^5-7y^2*z^4+6y^3*z^3+18y^4*z^2+8y^5*z+y^6)>0

在x<y<z条件下，上式中括号内易证大于零。所以只需证
z^6-2y*z^5-7y^2*z^4+6y^3*z^3+18y^4*z^2+8y^5*z+y^6≥0
<==> (z^3-yz^2-4zy^2-y^3)^2≥0

故不等式(1)得证，
不等式(1)下界系数7/5为最佳。