lasha wrote: Let $ x,y,z$ be positive real numbers,satisfying equality $ x^{2} + y^{2} + z^{2} = 25$. Find the minimal possible value of the expression $ \frac {xy}{z} + \frac {yz}{x} + \frac {zx}{y}$. We have : $ M^2=( \frac {xy}{z} + \frac {yz}{x} + \frac {zx}{y})^{2} \geq 3(x^2+y^2+z^2)=75$ $ Min P=5\sqrt{3} \Leftrightarrow x=y=z=\frac{5}{\sqrt{3}}$

But how do you know that $ \frac{5}{\sqrt3}$ ? AM GM will not always give you the minimum rite? BY AM GM, $ (a+b+c)^2 \ 3(ab+bc+ca)$ for postive reals $ a,b,c$. I know that this proof can be given using Cauchy but how do you know when one can get the minimum value of the expression?

I don't understand what's false in the solution. The author mentions the equality cases as well and only uses Cauchy inequality, which you wrote. (Namely $ (a+b+c)^{2}>=3(ab+bc+ca)$).

Yaaeehh... Easy one. Just see that \[ \frac{1}{3}\left( {\frac{{xy}}{z} + \frac{{yz}}{x} + \frac{{xz}}{y}} \right) \ge \sqrt {\frac{{x^2 + y^2 + ^2 }}{3}} = \sqrt {\frac{{25}}{3}} \,\,\,\,\, \] which gives us minimum value \[ \frac{{xy}}{z} + \frac{{yz}}{x} + \frac{{xz}}{y} \ge 5\sqrt 3 \]

The equation $ x^{2}+y^{2}+z^{2}=25$ is symmetric towards x,y,z. Also, $ \frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}$ is symmetric, therefore: $ \frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}=3\frac{xy}{z}$. Minumum value of the hemisphere is when x=y=z, therefore $ 3x^2=25$, $ x=\frac{5}{\sqrt{3}}$, which makes final answer $ \frac{5}{\sqrt{3}}*3=5\sqrt{3}$.

lagrange multiplier $F(x,y,z,L)=\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}+L(x^2+y^2+z^2-25)$ by derivatives we obtain $\frac{x^2y^2+x^2z^2-y^2z^2}{x^3yz}=\frac{x^2y^2+y^2z^2-x^2z^2}{xy^3z}=\frac{x^2z^2+y^2z^2-x^2y^2}{xyz^3}=-2L$ hence$(x^2-y^2)(x^2z^2+y^2z^2-x^2y^2)=0$ let $x^2=a,y^2=b,z^2=c$ then $a=b or (a+b)c=ab$ $b=c or (b+c)a=bc$ $c=a or (c+a)b=ca$ if at least two of the equations above are the latter one,by generality let $(a+b)c=ab,(b+c)a=bc$ then$bc>ab>bc$contradiction! so at least two are the former one,hencde $a=b=c=\frac{25}{3}$,yielding minimum $5\sqrt{3}$.

suppose $x\geq y\geq z$ \[ \frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y}= \sum_{cyc} x^2.\frac{y}{xz} \geq \frac{1}{3} (x^2+y^2+z^2)(\sum_{cyc} \frac{y}{xz}) \] \[ \Rightarrow \frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} \geq \frac{25^2}{3xyz} \geq 5\sqrt{3} \] equality for $x=y=z=\frac{5}{\sqrt{3}}$

lasha wrote: Let $ x,y,z$ be positive real numbers,satisfying equality $ x^{2}+y^{2}+z^{2}=25$. Find the minimal possible value of the expression $ \frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y}$. Prove that $( \frac {xy}{z} + \frac {yz}{x} + \frac {zx}{y})^{2} \geq 3(x^2+y^2+z^2) = 75$ . If we do that, the answer will be $\sqrt75$ which realises for $x=y=z=\frac {5}{\sqrt3}$. $( \frac {xy}{z} + \frac {yz}{x} + \frac {zx}{y})^{2}=\frac {x^2y^2}{z^2} + \frac {y^2z^2}{x^2} + \frac {z^2x^2}{y^2} + 2(x^2+y^2+z^2)$ $2(x^2+y^2+z^2)$ will be cancelled and obtained inequality will be: $\frac {x^2y^2}{z^2} + \frac {y^2z^2}{x^2} + \frac {z^2x^2}{y^2} \geq x^2+y^2+z^2$ With the simplyficiation of that expression: $\frac {\sum_{cyc} x^4y^4}{2} \geq \frac {\sum_{cyc} x^4y^2z^2}{2}$ This is right with Muirhead's inequality.

Alexander333 wrote: lasha wrote: Let $ x,y,z$ be positive real numbers,satisfying equality $ x^{2}+y^{2}+z^{2}=25$. Find the minimal possible value of the expression $ \frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y}$. Prove that $( \frac {xy}{z} + \frac {yz}{x} + \frac {zx}{y})^{2} \geq 3(x^2+y^2+z^2) = 75$ . If we do that, the answer will be $\sqrt75$ which realises for $x=y=z=\frac {5}{\sqrt3}$. $( \frac {xy}{z} + \frac {yz}{x} + \frac {zx}{y})^{2}=\frac {x^2y^2}{z^2} + \frac {y^2z^2}{x^2} + \frac {z^2x^2}{y^2} + 2(x^2+y^2+z^2)$ $2(x^2+y^2+z^2)$ will be cancelled and obtained inequality will be: $\frac {x^2y^2}{z^2} + \frac {y^2z^2}{x^2} + \frac {z^2x^2}{y^2} \geq x^2+y^2+z^2$ With the simplyficiation of that expression: $\frac {\sum_{cyc} x^4y^4}{2} \geq \frac {\sum_{cyc} x^4y^2z^2}{2}$ This is right with Muirhead's inequality. OR AM-GM $x^4y^4+y^4z^4\geq{2x^2y^4z^2}$....

SRIDARA wrote: OR AM-GM $x^4y^4+y^4z^4\geq{2x^2y^4z^2}$.... Muirhead's inequality and AM-GM are same yeah? or not?

YES!,but more claire!

lasha wrote: Let $ x,y,z$ be positive real numbers,satisfying equality $ x^{2}+y^{2}+z^{2}=25$. Find the minimal possible value of the expression $ \frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y}$. Greece 2004 juniors https://artofproblemsolving.com/community/c6h3419400p32930347