Inequality for two-variable functions Let $y\geq x\geq 1$ and $k \in \mathbb{N} \geq 1$. I want to find $f_{x,y} \geq \dfrac{1}{x^k}$ such that:

*

*$f(x,x) = \dfrac{1}{x^k}$ (This condition is to avoid trivial answers like $ y^2 \geq \dfrac{1}{x}$ for instance.)

This condition gives the general form for $f(x,y)$ which is $\sum a_i x^{b_i} y^{c_i} $ such that $b_i+c_i=-k$.


*$\sum$ $(\text{coefficients} \cdot (-\text{powers of } x)) < k$
If we took this form $f(x,y)=\sum a_i x^{b_i} y^{c_i}$ then the condition becomes $ \sum a_i · (-b_i) < k$.
For instance with $k=4$ and $f(x,y) = \dfrac{y}{x^5} \geq \dfrac{1}{x^4}$, the first condition holds but the second doesn't because $\text{coefficients}=1$ and $\text{power} = 5$ and so the total sum is $1\cdot 5= 5 >4$.
To find the sum in condition 2 for $f(x,y)= \dfrac{1}{x^2 y^2}+\dfrac{2}{x^3 y}$, we calculate $ 1\cdot 2+2\cdot 3 = 8 >4$.
I conjecture that there is no such function but I couldn't prove it.
 A: 
$\def\N{\mathbb{N}}\def\R{\mathbb{R}}\def\paren#1{\left(#1\right)}$Lemma 1: Suppose $a_1, \cdots, a_n, u_1, \cdots, u_n \in \R$ and $\sum\limits_{k = 1}^n a_k = 1$. If$$
\sum_{k = 1}^n a_k t^{u_k} \geqslant 1,\quad \forall 0 < t \leqslant 1
$$
then $\sum\limits_{k = 1}^n a_k u_k \leqslant 0$.

Proof: Define $F(t) = \sum\limits_{k = 1}^n a_k t^{u_k}$. Note that $F'(1)$ exists and $F(1) = \sum\limits_{k = 1}^n a_k = 1$. Suppose $F'(1) > 0$, there exists $0 < δ < 1$ such that $F(t) < F(1) = 1$ for $t \in (1 - δ, 1)$, a contradiction. Thus $F'(1) \leqslant 0$, i.e.\begin{gather*}
F'(1) = \left.\paren{ \sum_{k = 1}^n a_k u_k t^{u_k - 1} }\right|_{t = 1} = \sum_{k = 1}^n a_k u_k \leqslant 0. \tag*{$\Box$}
\end{gather*}
Now return to the question. The proposition to be proved can be formalized as below (@RiverLi):

Proposition: Suppose $m, n \in \N_+$. For any $a_1, \cdots, a_n, b_1, \cdots, b_n, c_1, \cdots, c_n \in \R$, if $\sum\limits_{k = 1}^n a_k = 1$, $b_k + c_k = -m$ for $1 \leqslant k \leqslant n$, and\begin{gather*}
\sum_{k = 1}^n a_k x^{b_k} y^{c_k} \geqslant \frac{1}{x^m},\quad \forall y \geqslant x \geqslant 1 \tag{1}
\end{gather*}
then $\sum\limits_{k = 1}^n a_k · (-b_k) \geqslant m$.

Proof: Since $b_k + c_k = -m$ for all $k$, then$$
x^m \sum_{k = 1}^n a_k x^{b_k} y^{c_k} = \sum_{k = 1}^n a_k x^{b_k + m} y^{c_k} = \sum_{k = 1}^n a_k \paren{ \frac{x}{y} }^{b_k + m},
$$
thus (1) is equivalent to$$
\sum_{k = 1}^n a_k t^{b_k + m} \geqslant 1.\quad \forall 0 < t \leqslant 1
$$
Therefore, lemma 1 implies that $\sum\limits_{k = 1}^n a_k (b_k + m) \leqslant 0$, then\begin{gather*}
-\sum_{k = 1}^n a_k b_k \geqslant m \sum_{k = 1}^n a_k = m. \tag*{$\Box$}
\end{gather*}

In fact, the condition in lemma 1 is also necessary when $a_1, \cdots, a_n > 0$.

Lemma 2: Suppose $a_1, \cdots, a_n > 0$, $u_1, \cdots, u_n \in \R$. If $\sum\limits_{k = 1}^n a_k = 1$ and $\sum\limits_{k = 1}^n a_k u_k \leqslant 0$, then$$
\sum_{k = 1}^n a_k t^{u_k} \geqslant 1.\quad \forall 0 < t \leqslant 1
$$

Proof: Because $\sum\limits_{k = 1}^n a_k = 1$, the AM-GM inequality implies that for $0 < t \leqslant 1$,\begin{gather*}
\sum_{k = 1}^n a_k t^{u_k} \geqslant \prod_{k = 1}^n (t^{u_k})^{a_k} = t^{\sum\limits_{k = 1}^n a_k u_k} \geqslant 1. \tag*{$\Box$}
\end{gather*}
