If `a, b gt 0`, then the least value of `(a^(2) - ab + b^(2))/((a + b)^(2))` is ______

Answer

To find the least value of the expression \(\frac{a^2 - ab + b^2}{(a + b)^2}\) for \(a, b > 0\), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ \frac{a^2 - ab + b^2}{(a + b)^2} \]

### Step 2: Simplify the numerator
Notice that we can rewrite the numerator \(a^2 - ab + b^2\) as follows:
\[ a^2 - ab + b^2 = a^2 + b^2 - ab \]
We will also use the identity \(a^2 + b^2 = (a + b)^2 - 2ab\). Thus, we have:
\[ a^2 - ab + b^2 = (a + b)^2 - 2ab - ab = (a + b)^2 - 3ab \]

### Step 3: Substitute back into the expression
Now substituting back into our original expression gives:
\[ \frac{(a + b)^2 - 3ab}{(a + b)^2} \]

### Step 4: Split the fraction
This can be split into two parts:
\[ 1 - \frac{3ab}{(a + b)^2} \]

### Step 5: Analyze the term \(\frac{ab}{(a + b)^2}\)
Let \(x = \frac{a}{b}\). Then we can express \(ab\) in terms of \(x\) and \(b\):
\[ ab = b^2 \cdot x \]
Thus, we can rewrite \(\frac{ab}{(a + b)^2}\) as:
\[ \frac{ab}{(a + b)^2} = \frac{bx}{(b(x + 1))^2} = \frac{x}{b(x + 1)^2} \]

### Step 6: Use AM-GM inequality
By the AM-GM inequality, we know:
\[ \frac{a + b}{2} \geq \sqrt{ab} \implies \left(\frac{a + b}{2}\right)^2 \geq ab \implies \frac{(a + b)^2}{4} \geq ab \]
Thus, we have:
\[ \frac{ab}{(a + b)^2} \leq \frac{1}{4} \]

### Step 7: Substitute back into the expression
Substituting this back, we find:
\[ 1 - 3 \cdot \frac{ab}{(a + b)^2} \geq 1 - 3 \cdot \frac{1}{4} = 1 - \frac{3}{4} = \frac{1}{4} \]

### Step 8: Conclusion
Thus, the least value of the expression \(\frac{a^2 - ab + b^2}{(a + b)^2}\) is:
\[ \frac{1}{4} \]

### Final Answer
The least value of \(\frac{a^2 - ab + b^2}{(a + b)^2}\) is \(\frac{1}{4}\).

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