If a, b, c are positive numbers, then prove that (1+a)^7(1+b)^7(1+c)^7>7^7 a^4 b^4 c^4

n A.M. >= G.M. n (a+b+c+a b+b c+c a+a b c / 7) >= (a^4 b^4 c^4)^(1 / 7) n a+b+c+a b+b c+c a+a b c >= 7(a^4 b^4 c^4)^1 / 7 n => 1+a+b+c+a b+b c+c a+a b c >= 7(a^4 b^4 c^4)^1 / 7 n => (1+a)(1+b)(1+c) >= 7(a^4 b^4 c^4)^1 / 7 n => (1+a)^7(1+b)^7(1+c)^7 >= 7^7(a^4 b^4 c^4)n\

If a, b, c are positive numbers, then prove that (1+a)^... | Sequence & Series (Mathematics)

\begin{aligned}\n& \text { A.M. } \geq \text { G.M. } \\\n& \frac{a+b+c+a b+b c+c a+a b c}{7} \geq\left(a^4 b^4 c^4\right)^{\frac{1}{7}} \\\n& a+b+c+a b+b c+c a+a b c \geq 7\left(a^4 b^4 c^4\right)^{1 / 7} \\\n& \Rightarrow 1+a+b+c+a b+b c+c a+a b c \geq 7\left(a^4 b^4 c^4\right)^{1 / 7} \\\n& \Rightarrow(1+a)(1+b)(1+c) \geq 7\left(a^4 b^4 c^4\right)^{1 / 7} \\\n& \Rightarrow(1+a)^7(1+b)^7(1+c)^7 \geq 7^7\left(a^4 b^4 c^4\right)\n\end{aligned}

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

Let a_{1},a_{2},a_{3},. be a G.P. of increasing positive numbers. Let the sum of its 6^{{th}{~}} and 8^{{th}{~}} terms be 2 and the product of its 3^{{rd}{~}} and 5^{{th}{~}} terms be 1/9. Then 6( a_{2} + a_{4} )( a_{4} + a_{6} ) is equal to

For any odd integer n ? 1,n^{3} - (n - 1)^{3} + + ( - 1)^{n - 1}1^{3} =

The sum of the first n terms of the series (1^{2} + 2.2^{2} + 3^{2} + 2.4^{2} + 5^{2} + 2 . 6^{2} + ..) is (\frac{n(n + 1)^{2}}{2}, when n is even. when n is odd, the sum is

Let a,b be two non-zero real numbers. If p and r are the roots of the equation x^{2} - 8ax + 2a = 0 and q and s are the roots of the equation \end{enumerate} x^{2} + 12bx + 6b = 0, such that \frac{1}{p},\frac{1}{q},\frac{1}{r},\frac{1}{s} are in A.P., then a^{- 1} - b^{- 1} is equal to

If a_{1},a_{2},a_{3}, are in A.P. such that a_{1} + a_{7} + a_{16} = 40, then the sum of the first 15 terms of this A.P. is

Let \frac{1}{x_{1}},\frac{1}{x_{2}},,\frac{1}{x_{n}}( x_{i} eq 0 .\ for . \ i = 1,2,,n ) be in A.P. such that x_{1} = 4 and x_{21} = 20. If n is the least positive integer for which x_{n} > 50, then \sum_{i = 1}^{n}\mspace{2mu}( \frac{1}{x_{i}} ) is equal to

Let 3,a,b,c be in A.P. and 3,a - 1,b + 1,c + 9 be in G.P. Then, the arithmetic mean of a,b and c is

Number of 4 -digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11 , is equal to

Download Divine JEE App

Quick Links

Exam Categories

Contact Info

Divine JEE