This is not quite the form you requested. But if your goal was to control chi-squared left tails, I have found these lower bounds helpful.

For $X_1 \sim \chi^2_1$:

Let $Z \sim N(0,1)$, so that $Z/\sqrt{2} \sim N(0, 1/2)$. For $x>0$, we have $\text{erf}(x) = P\left(|Z/\sqrt{2}| \leq x\right) = P(X_1 \leq 2x)$.

Apply this bound or this refinement: $\text{erf}(x) \leq \sqrt{1-\exp(-4x^2/\pi)}$.

Then $P(X_1 \geq y) \geq 1- \sqrt{1-\exp(-2y/\pi)}$ for $y>0$.

For $X_2 \sim \chi^2_2$:

Directly from the chi-squared CDF for $n=2$, we see that $P(X_2 \geq x) = e^{-x/2}$.

Also, since the chi-squared distribution shifts right as the df increases, $P(X_n \geq x) \geq e^{-x/2}$ for $n\geq 2$.

For other $n$:

See this answer to another question.

This is not quite the form you requested. But if your goal was to control chi-squared left tails, I have found these lower bounds helpful.

For $X_1 \sim \chi^2_1$:

Let $Z \sim N(0,1)$, so that $Z/\sqrt{2} \sim N(0, 1/2)$. For $x>0$, we have $\text{erf}(x) = P\left(|Z/\sqrt{2}| \leq x\right) = P(X_1 \leq 2x)$.

Apply this bound or this refinement: $\text{erf}(x) \leq \sqrt{1-\exp(-4x^2/\pi)}$.

Then $P(X_1 \geq y) \geq 1- \sqrt{1-\exp(-2y/\pi)}$ for $y>0$.

For $X_2 \sim \chi^2_2$:

Directly from the chi-squared CDF for $n=2$, we see that $P(X_2 \geq x) = e^{-x/2}$.

Also, since the chi-squared distribution shifts right as the df increases, $P(X_n \geq x) \geq e^{-x/2}$ for $n\geq 2$.

For other $n$:

See this answer to another question.

This is not quite the form you requested. But if your goal was to control chi-squared left tails, I have found these lower bounds helpful.

For $X_1 \sim \chi^2_1$:

Let $Z \sim N(0,1)$, so that $Z/\sqrt{2} \sim N(0, 1/2)$. For $x>0$, we have $\text{erf}(x) = P\left(|Z/\sqrt{2}| \leq x\right) = P(X_1 \leq 2x)$.

Apply this bound or this refinement: $\text{erf}(x) \leq \sqrt{1-\exp(-4x^2/\pi)}$.

Then $P(X_1 \geq y) \geq 1- \sqrt{1-\exp(-2y/\pi)}$ for $y>0$.

For $X_2 \sim \chi^2_2$:

Directly from the chi-squared CDF for $n=2$, we see that $P(X_2 \geq x) = e^{-x/2}$.

Also, since the chi-squared distribution shifts right as the df increases, $P(X_n \geq x) \geq e^{-x/2}$ for $n\geq 2$.

For other $n$:

See this answer to another question.

This is not quite the form you requested. But if your goal was to control chi-squared left tails, I have found these lower bounds helpful.

For $X_1 \sim \chi^2_1$:

Let $Z \sim N(0,1)$, so that $Z/\sqrt{2} \sim N(0, 1/2)$. For $x>0$, we have $\text{erf}(x) = P\left(|Z/\sqrt{2}| \leq x\right) = P(X_1 \leq 2x)$.

Apply this bound or this refinement: $\text{erf}(x) \leq \sqrt{1-\exp(-4x^2/\pi)}$.

Then $P(X_1 \geq y) \geq 1- \sqrt{1-\exp(-2y/\pi)}$ for $y>0$.

For $X_2 \sim \chi^2_2$:

Directly from the chi-squared CDF for $n=2$, we see that $P(X_2 \geq x) = e^{-x/2}$.

Also, since the chi-squared distribution shifts right as the df increases, $P(X_n \geq x) \geq e^{-x/2}$ for $n\geq 2$.

For other $n$:

See this answer to another question.