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Algebra Difficulty 6.6 National olympiad Prove it

51. Given that x,y,zx, y, z are positive numbers, prove that xy2+z2+yz2+x2+zx2+y2>2\frac{x}{\sqrt{y^{2}+z^{2}}}+\frac{y}{\sqrt{z^{2}+x^{2}}}+\frac{z}{\sqrt{x^{2}+y^{2}}}>2. (2005 Macau Mathematical Olympiad Problem)

Solution

51. By Hölder's inequality, we have
(xy2+z2+yz2+x2+zx2+y2)2[x(y2+z2)+y(z2+x2)+z(x2+y2)](x+y+z)3\begin{array}{l} \left(\frac{x}{\sqrt{y^{2}+z^{2}}}+\frac{y}{\sqrt{z^{2}+x^{2}}}+\frac{z}{\sqrt{x^{2}+y^{2}}}\right)^{2}\left[x\left(y^{2}+z^{2}\right)+y\left(z^{2}+x^{2}\right)+z\left(x^{2}+y^{2}\right)\right] \geqslant \\ (x+y+z)^{3} \end{array}

It suffices to prove that
(x+y+z)3>4[x(y2+z2)+y(z2+x2)+z(x2+y2)]=4(x+y+z)(xy+yz+zx)12xyz\begin{aligned} (x+y+z)^{3}> & 4\left[x\left(y^{2}+z^{2}\right)+y\left(z^{2}+x^{2}\right)+z\left(x^{2}+y^{2}\right)\right]= \\ & 4(x+y+z)(x y+y z+z x)-12 x y z \end{aligned}

By Schur's inequality, (x+y+z)34(x+y+z)(xy+yz+zx)+9xyz0(x+y+z)^{3}-4(x+y+z)(x y+y z+z x)+9 x y z \geqslant 0, so
(x+y+z)34(x+y+z)(xy+yz+zx)9xyz>4(x+y+z)(xy+yz+zx)12xyz\begin{array}{l} (x+y+z)^{3} \geqslant 4(x+y+z)(x y+y z+z x)-9 x y z> \\ 4(x+y+z)(x y+y z+z x)-12 x y z \end{array}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.