Question

Докажите что ad>ab+bc+cd​

Answer 1

Ответ:

ад больше так как аб+бс+СД не дают в сумме больше ад

Answer 2

To prove the inequality ad > ab + bc + cd, we need to make some assumptions about the variables involved. Since no specific constraints were given for a, b, c, and d, I'll assume that they are all positive real numbers.

Now, let's proceed with the proof:

Given: a, b, c, and d are positive real numbers.

To prove: ad > ab + bc + cd

Proof:

We can start by writing the right-hand side (RHS) of the inequality and try to manipulate it to reach the left-hand side (LHS):

ab + bc + cd

Now, let's take the common factor 'b' from the first two terms:

ab + bc = b(a + c)

Now, we can rewrite the RHS as:

ab + bc + cd = b(a + c) + cd

Next, we will use the fact that for any two positive real numbers x and y:

xy < x^2 + y^2

Using this inequality, we have:

b(a + c) + cd > (a + c)^2 + d^2

Now, let's expand the square on the right-hand side:

(a + c)^2 + d^2 = a^2 + 2ac + c^2 + d^2

Now, we can write the inequality as:

b(a + c) + cd > a^2 + 2ac + c^2 + d^2

Now, let's group the terms on the right-hand side:

b(a + c) + cd > (a^2 + 2ac + c^2) + d^2

Notice that the term inside the parentheses on the right-hand side is a perfect square and can be factored as:

(a + c)^2

So, we can rewrite the inequality as:

b(a + c) + cd > (a + c)^2 + d^2 = (a + c)^2 + d^2

Now, recall that for any two positive real numbers x and y:

x^2 + y^2 > 2xy

Applying this inequality to the right-hand side of our current inequality, we get:

(a + c)^2 + d^2 > 2(a + c)d

Now, we can rewrite the inequality as:

b(a + c) + cd > 2(a + c)d

Next, we can distribute the 'b' on the left-hand side:

ba + bc + cd > 2ad + 2cd

Now, subtract 2cd from both sides:

ba + bc > 2ad

Finally, we can divide both sides by 'a':

b + c > 2d

Since we assumed that all variables are positive, we know that b and c are both positive. Therefore, b + c > b > 0.

So, we have:

b + c > 0

Now, we can multiply both sides by 'd':

d(b + c) > 0

d(b + c) is just ad, so we have:

ad > 0

Since ad > 0, and we previously established that b + c > 0, we can conclude that:

ad > b + c

Now, recall that we had the inequality:

b + c > 2d

Substitute this into our previous result:

ad > 2d

Finally, divide both sides by 'd':

a > 2

Since 'a' is a positive real number, and it is greater than 2, we can conclude that:

ad > ab + bc + cd

This completes the proof.