Question

Довести (m^2+n^2)/2>=m+n-1

Answer 1

m^2+n^2-2m-2n+2=(m^2-2m+1)-1+(n^2-2n+1)-1+2=(m-1)^2+(n-1)^2

(m-1)^2>=0,(n-1)^2>=0,тогда (m-1)^2+(n-1)^2>=0.

Answer 2

To prove the inequality $\frac{{m^2 + n^2}}{2} \geq m + n - 1$, we'll use algebraic manipulation. Let's go step by step:

Step 1: Expand the left-hand side (LHS) of the inequality: $\frac{{m^2 + n^2}}{2} = \frac{1}{2}m^2 + \frac{1}{2}n^2$.

Step 2: Now, we need to prove $\frac{1}{2}m^2 + \frac{1}{2}n^2 \geq m + n - 1$.

Step 3: Subtract $m$ from both sides of the inequality: $\frac{1}{2}m^2 + \frac{1}{2}n^2 - m \geq n - 1$.

Step 4: Subtract $n$ from both sides of the inequality: $\frac{1}{2}m^2 + \frac{1}{2}n^2 - m - n \geq -1$.

Step 5: Bring all the terms to the left-hand side: $\frac{1}{2}m^2 - m + \frac{1}{2}n^2 - n + 1 \geq 0$.

Step 6: Now, we want to prove the expression on the left-hand side $\frac{1}{2}m^2 - m + \frac{1}{2}n^2 - n + 1$ is greater than or equal to zero for all real values of $m$ and $n$.

Step 7: Notice that the left-hand side can be factored as follows: $\frac{1}{2}m^2 - m + \frac{1}{2}n^2 - n + 1 = \frac{1}{2}(m^2 - 2m + 1) + \frac{1}{2}(n^2 - 2n + 1)$.

Step 8: Simplify the squares and constants within the parentheses: $\frac{1}{2}(m - 1)^2 + \frac{1}{2}(n - 1)^2$.

Step 9: Since squares are always non-negative, both terms $\frac{1}{2}(m - 1)^2$ and $\frac{1}{2}(n - 1)^2$ are greater than or equal to zero.

Step 10: Therefore, the sum of these non-negative terms is also greater than or equal to zero: $\frac{1}{2}(m - 1)^2 + \frac{1}{2}(n - 1)^2 \geq 0$.

Step 11: This implies that $\frac{1}{2}m^2 - m + \frac{1}{2}n^2 - n + 1 \geq 0$.

Step 12: And since we have proved this for all real values of $m$ and $n$, we have also proved the original inequality $\frac{{m^2 + n^2}}{2} \geq m + n - 1$