Question

Find all pairs ( x , y ) of real numbers such that -


$16^ { x^2 + y } + 16^ { x + y^2 } = 1$ ​

Answer 1

Answer :

We know the application of inequality,

In which, a system of linear inequalities is used to determine the maximum or minimum values of a situation with multiple constraints.

So, We will use application of inequality here.

By application of inequality,

$\implies \sf a+b\geq 2ab$

Now, we have :

$\sf \implies 16^{\largetext{x^2 + y}}+\;16^{\largetext{x + y}}2\geq 2\sqrt{16^{\largetext{x^2 + y}}\times16^{x + y2}}$

$\sf \implies 2\times 4^{x^2+x+y^2+y}$

So,

$\implies \sf t^2+t\geq -14\;clench\;for\;all\;t\;\epsilon\;R\;and\;for\;all\;x,y\;\epsilon\;R,\\ \\ \\ \textbf {We\;have,}\\ \\ 16^{x^2+y}+16^{x+y^2}\geq 2\times 4^{x^2+x+y^2+y}\geq 2\times4-^{\frac{1}{2}}=1$

Hence,

With equality in case x = y = $\sf \dfrac{1}{2}$

Answer 2

Answer:

Answer :

We know the application of inequality,

In which, a system of linear inequalities is used to determine the maximum or minimum values of a situation with multiple constraints.

So, We will use application of inequality here.

By application of inequality,

\implies \sf a+b\geq 2ab⟹a+b≥2ab

Now, we have :

$$\begin{gathered}\sf \implies 16^{\largetext{x^2 + y}}+\;16^{\largetext{x + y}}2\geq 2\sqrt{16^{\largetext{x^2 + y}}\times16^{x + y2}}\\ \\ \\\end{gathered}$$

$$\sf \implies 2\times 4^{x^2+x+y^2+y}$$

So,

$$\begin{gathered}\implies \sf t^2+t\geq -14\;clench\;for\;all\;t\;\epsilon\;R\;and\;for\;all\;x,y\;\epsilon\;R,\\ \\ \\ \textbf {We\;have,}\\ \\ 16^{x^2+y}+16^{x+y^2}\geq 2\times 4^{x^2+x+y^2+y}\geq 2\times4-^{\frac{1}{2}}=1\end{gathered}$$

Hence,

With equality in case x = y = $$\sf \dfrac{1}{2}$$