x,y,z正整数 x>y>z证明 x^2x +y^2y+z^2z>x^(y+z)*y^(x+z)*z^(x+y)x,y,z正整数 x>y>z证明 x^2x * y^2y * z^2z>x^(y+z)*y^(x+z)*z^(x+y)不是+是 *
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x,y,z正整数x>y>z证明x^2x+y^2y+z^2z>x^(y+z)*y^(x+z)*z^(x+y)x,y,z正整数x>y>z证明x^2x*y^2y*z^2z>x^(y+z)*y^(x+z)*z
x,y,z正整数 x>y>z证明 x^2x +y^2y+z^2z>x^(y+z)*y^(x+z)*z^(x+y)x,y,z正整数 x>y>z证明 x^2x * y^2y * z^2z>x^(y+z)*y^(x+z)*z^(x+y)不是+是 *
x,y,z正整数 x>y>z证明 x^2x +y^2y+z^2z>x^(y+z)*y^(x+z)*z^(x+y)
x,y,z正整数 x>y>z证明 x^2x * y^2y * z^2z>x^(y+z)*y^(x+z)*z^(x+y)
不是+是 *
x,y,z正整数 x>y>z证明 x^2x +y^2y+z^2z>x^(y+z)*y^(x+z)*z^(x+y)x,y,z正整数 x>y>z证明 x^2x * y^2y * z^2z>x^(y+z)*y^(x+z)*z^(x+y)不是+是 *
正整数?取对数
即证:2xlnx+2ylny+2zlnz>(y+z)lnx+(x+z)lny+(x+y)lnz
x>y>z,lnx>lny>lnz
由排序不等式得
xlnx+ylny+zlnz>ylnx+zlny+xlnz
xlnx+ylny+zlnz>zlnx+xlny+ylnz
两式相加,即得要证的式子
研究员
x,y,z正整数 x>y>z证明 x^2x +y^2y+z^2z>x^(y+z)*y^(x+z)*z^(x+y)x,y,z正整数 x>y>z证明 x^2x * y^2y * z^2z>x^(y+z)*y^(x+z)*z^(x+y)不是+是 *
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