If `a+b+c+d=1` then maximum value of `ab+ bc + cd + da`

Answer

To find the maximum value of \( ab + bc + cd + da \) given that \( a + b + c + d = 1 \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression we want to maximize:
\[ ab + bc + cd + da \]
We can group the terms:
\[ ab + da + bc + cd = b(a + c) + d(c + a) \]

### Step 2: Use the constraint
Since we know that \( a + b + c + d = 1 \), we can express \( d \) in terms of \( a, b, \) and \( c \):
\[ d = 1 - (a + b + c) \]
This allows us to rewrite our expression in terms of \( a, b, \) and \( c \).

### Step 3: Apply the Arithmetic Mean-Geometric Mean Inequality (AM-GM)
To find the maximum value, we can apply the AM-GM inequality. We know that:
\[ \frac{a + b + c + d}{4} \geq \sqrt[4]{abcd} \]
Since \( a + b + c + d = 1 \), we have:
\[ \frac{1}{4} \geq \sqrt[4]{abcd} \]
This implies:
\[ abcd \leq \left(\frac{1}{4}\right)^4 = \frac{1}{256} \]

### Step 4: Substitute and simplify
Now, we can try to express \( ab + bc + cd + da \) in terms of \( a + c \) and \( b + d \):
Let \( x = a + c \) and \( y = b + d \). Then:
\[ x + y = 1 \]
We want to maximize:
\[ ab + cd + da + bc = xy \]
Using the AM-GM inequality again:
\[ xy \leq \left(\frac{x + y}{2}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

### Step 5: Conclusion
Thus, the maximum value of \( ab + bc + cd + da \) is:
\[ \boxed{\frac{1}{4}} \]