<s>[INST] Show that $x^2 + y^2 + z^2 \ge 35$ if $x+3y+5z \ge 35.$ Show that $x^2 + y^2 + z^2 \ge 35$ if $x+3y+5z \ge 35.$
I have tried everything (proof by contradiction, etc.) but I can't seem to get it. The book didn't give any constraints whatsoever. Any hints would be appreciated. Thank you.
 [/INST] I suggest a solution for who know Lagrange multiplier.
Let $f(x,y,z)=x^2+y^2+z^2$, and $g(x,y,z)=x+3y+5z=a\ge 35$. Then
\begin{align}
\nabla f(x,y,z)&=(2x,2y,2z)\\
\nabla g(x,y,z)&=(1,3,5)\ne \mathbf{0}
\end{align}
Let $\lambda$ be a real number such that $\nabla f = \lambda \nabla g$, then
$$
2x=\lambda,\quad 2y=3\lambda,\quad 2z=5\lambda.
$$
Thus
$$
x+3y+5z=\frac{\lambda}{2}+\frac{9\lambda}{2}+\frac{25\lambda}{2}=\frac{35\lambda}{2},
$$
so $\lambda=\frac{2a}{35}$. Therefore,
\begin{align}
x^2+y^2+z^2&=\frac{\lambda^2}{4}+\frac{9\lambda^2}{4}+\frac{25\lambda^2}{4}\\
&= \frac{35}{4}\cdot \lambda^2\\
&=\frac{35}{4}\cdot \frac{4a^2}{35^2}\\
&=\frac{a^2}{35}\\
&\ge 35.
\end{align}
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