# Prove Inequality x,y with $x+y=2$

• MHB
• maxkor
Finally, we can use the fact that $x,y>0$ to show that $\sqrt[3]{(x^2+7)(y^2+7)}\geq\sqrt{xy}+\sqrt{7}$, which can be proven using the AM-GM inequality.Putting everything together, we have:$2(4+\sqrt[3]{(x^2+7)(y^2+7)})\geq2(\sqrt maxkor Let x,y>0 and x+y=2. Prove$\sqrt{x+\sqrt[3]{y^2+7}}+\sqrt{y+\sqrt[3]{x^2+7}}\geq2\sqrt3$Hello! To prove this inequality, we can use the Cauchy-Schwarz inequality, which states that for any real numbers$a_1, a_2, ..., a_n$and$b_1, b_2, ..., b_n$,$(a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2)\geq(a_1b_1+a_2b_2+...+a_nb_n)^2$In our case, we can let$a_1=\sqrt{x}$,$a_2=\sqrt[3]{y^2+7}$,$b_1=1$, and$b_2=1$. Then, the Cauchy-Schwarz inequality becomes:$(x+1)(\sqrt[3]{y^2+7}+1)\geq(\sqrt{x}+\sqrt[3]{y^2+7})^2$Similarly, we can also let$a_1=\sqrt{y}$,$a_2=\sqrt[3]{x^2+7}$,$b_1=1$, and$b_2=1$. The Cauchy-Schwarz inequality then becomes:$(y+1)(\sqrt[3]{x^2+7}+1)\geq(\sqrt{y}+\sqrt[3]{x^2+7})^2$Now, we can add these two inequalities together and use the fact that$x+y=2$to get:$(x+1)(\sqrt[3]{y^2+7}+1)+(y+1)(\sqrt[3]{x^2+7}+1)\geq(\sqrt{x}+\sqrt[3]{y^2+7})^2+(\sqrt{y}+\sqrt[3]{x^2+7})^2$Simplifying, we get:$2(x+y+2+\sqrt[3]{(x^2+7)(y^2+7)})\geq2(\sqrt{x}+\sqrt{y}+\sqrt[3]{x^2+7}+\sqrt[3]{y^2+7})^2$Using the fact that$x+y=2$, we can further simplify to:$2(4+\

## 1. How can I prove the inequality x < y when x + y = 2?

To prove this inequality, we can use the fact that if a + b = c, then a < c and b < c. So, in this case, if x + y = 2, then x < 2 and y < 2. Since x and y are both less than 2, x must be less than y.

## 2. Can I use algebra to prove this inequality?

Yes, you can use algebra to prove this inequality. One way to do so is by subtracting y from both sides of the equation x + y = 2, which gives us x = 2 - y. Then, we can substitute this value of x into the inequality x < y to get (2 - y) < y. By solving for y, we can see that this is true for all values of y that are less than 1.

## 3. Is there a graphical way to prove this inequality?

Yes, there is a graphical way to prove this inequality. We can plot the points (x, y) where x + y = 2 on a coordinate plane and then draw a line connecting these points. This line will represent all possible values of x and y that satisfy the equation. We can then shade the region below this line to represent the inequality x < y. This visually shows that all points below the line satisfy the inequality.

## 4. Can I prove this inequality by contradiction?

Yes, you can prove this inequality by contradiction. Assume that x ≥ y, and then add x and y to get 2x ≥ 2. Since x and y are both positive, this means that x ≥ 1. But if x ≥ 1, then y ≤ 1, which contradicts the fact that x + y = 2. Therefore, our assumption must be false, and x < y.

## 5. Is this inequality always true?

Yes, this inequality is always true when x and y are positive and x + y = 2. This can be seen by looking at the graph of x + y = 2 and the line representing x < y. The line will always be below the graph, showing that x < y is always true for this equation.

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