Question

Let $x,y\in R$, satisfies ${{x}^{2}}+{{y}^{2}}+xy=1$. Find the minimum value of ${{x}^{3}}y+x{{y}^{3}}+3$.

Answer 1

Hint: First we consider the equation given, that is ${{x}^{2}}+{{y}^{2}}+xy=1$. Then we divide the equation into two parts ${{x}^{2}}+{{y}^{2}},xy$. Then we use the property that the arithmetic mean is greater than the geometric mean of the two numbers. By applying that property to the above parts and simplifying it, we get the minimum value of ${{x}^{3}}y+x{{y}^{3}}$. Then we add 3 to the above obtained value to find the required value.

Complete step-by-step answer:
We are given that ${{x}^{2}}+{{y}^{2}}+xy=1$.
Now let us consider the property that the Arithmetic Mean is greater than the Geometric Mean.
Now let us divide ${{x}^{2}}+{{y}^{2}}+xy=1$ into two parts ${{x}^{2}}+{{y}^{2}},xy$. Applying the above property that arithmetic mean is greater than the geometric mean.
\[\begin{align}
  & \Rightarrow \dfrac{{{x}^{2}}+{{y}^{2}}+xy}{2}\ge \sqrt[2]{\left( {{x}^{2}}+{{y}^{2}} \right)\times xy} \\
 & \Rightarrow \dfrac{1}{2}\ge \sqrt[2]{\left( {{x}^{3}}y+x{{y}^{3}} \right)} \\
 & \Rightarrow {{\left( \dfrac{1}{2} \right)}^{2}}\ge {{x}^{3}}y+x{{y}^{3}} \\
 & \Rightarrow {{x}^{3}}y+x{{y}^{3}}\le \dfrac{1}{4} \\
\end{align}\]
So, we get that \[{{x}^{3}}y+x{{y}^{3}}\le \dfrac{1}{4}\].
Now, let us consider the equation ${{x}^{3}}y+x{{y}^{3}}+3$.
\[\begin{align}
  & \Rightarrow {{x}^{3}}y+x{{y}^{3}}+3\le \dfrac{1}{4}+3 \\
 & \Rightarrow {{x}^{3}}y+x{{y}^{3}}+3\le \dfrac{13}{4} \\
\end{align}\]
Hence the answer is \[\dfrac{13}{4}\].

Note: The common misconception that happens while solving this question is one might take the inequality wrong as the Arithmetic mean is less than the geometric mean of the two numbers.