parmenides51 wrote: Let $a, b,c$ be positive real numbers satisfying the condition $$(x + y + z) \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)= 10$$Find the greatest value and the least value of $$T = (x^2 + y^2 + z^2) \left(\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2}\right)$$Trần Nam Dũng https://artofproblemsolving.com/community/c6h1133750p9191647 https://artofproblemsolving.com/community/c6h1529197p9186186

For the maximum, let $p=x+y+z,\;q=xy+yz+zx$ and $r=xyz$. The given condition is just $pq=10r,$ so we substitute into $T$ and simplify, we get: \[T=(p^{2}-2q)(\frac{q^{2}}{r^{2}}-2\frac{p}{r})=20(7-\frac{p^{2}}{q}-\frac{10q}{p^{2}})\le _{AM-GM} 20(7-2\sqrt{10})=140-40\sqrt{10}\]which can be achieved at $(x,y,z)=(x_{0},\frac{1}{x_{0}},1),$ where $x_{0}+\frac{1}{x_{0}}=\sqrt{10}-1.$ So the maximum is $140-40\sqrt{10}.$ Now, for the minimum, let \[s=\frac{x}{y}+\frac{y}{z}+\frac{z}{x},\;\;t=\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\]Then the given condition is equivalent to $s+t=7,$ and we can see that \[T=3+\sum_{cyc}\frac{x^{2}}{y^{2}}+\sum_{cyc}\frac{y^{2}}{x^{2}}=3+(s^{2}-2t)+(t^{2}-2s)=s^{2}+t^{2}-11\ge _{Cauchy}\frac{(s+t)^{2}}{2}-11=\frac{27}{2}\]which is achievable at $(x,y,z)=(2,1,1).$ So the minimum is $\frac{27}{2}.$

Let $a,b,c$ are positive real numbers such that$\frac {a}{b+c}+\frac {b}{c+a}+\frac {c}{a+b}=\frac{5}{3} .$ Prove that $$10\leq (a+b+c)\left(\frac 1{a}+\frac 1{b}+\frac 1{c}\right)\leq \frac {35}{3} $$Let $a,b,c$ are positive real numbers such that$\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}=5 .$ Prove that $$10\leq (a+b+c)\left(\frac 1{a}+\frac 1{b}+\frac 1{c}\right)\leq 5+4\sqrt{2} $$

Let $a,b,c$ are positive real numbers such that$(x^2 + y^2 + z^2) \left(\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2}\right)=11 .$ Prove that $$ 5-2\sqrt{5} \leq(x + y + z) \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)\leq 5+2\sqrt{5}$$