Mixing
Proof of inequality using Cauchy–Schwarz inequality
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How I can prove this inequality? How can I prove it using Cauchy–Schwarz inequality?
Let $a_1,...,a_n,b_1,....,b_n$ are any real numbers.
$sqrt {sum_{k=1}^n (a_k+b_k)^2)} leq sqrt {sum_{k=1}^n {a_k}^2} sqrt {sum_{k=1}^n {b_k}^2}$
real-analysis analysis inequality
edited Jan 19 at 0:09
Bernard
121k740116
asked Jan 19 at 0:00
AlarAlar
1
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add a comment |
$begingroup$
How I can prove this inequality? How can I prove it using Cauchy–Schwarz inequality?
Let $a_1,...,a_n,b_1,....,b_n$ are any real numbers.
$sqrt {sum_{k=1}^n (a_k+b_k)^2)} leq sqrt {sum_{k=1}^n {a_k}^2} sqrt {sum_{k=1}^n {b_k}^2}$
real-analysis analysis inequality
edited Jan 19 at 0:09
Bernard
121k740116
asked Jan 19 at 0:00
AlarAlar
1
$endgroup$
add a comment |
$begingroup$
How I can prove this inequality? How can I prove it using Cauchy–Schwarz inequality?
Let $a_1,...,a_n,b_1,....,b_n$ are any real numbers.
$sqrt {sum_{k=1}^n (a_k+b_k)^2)} leq sqrt {sum_{k=1}^n {a_k}^2} sqrt {sum_{k=1}^n {b_k}^2}$
real-analysis analysis inequality
edited Jan 19 at 0:09
Bernard
121k740116
asked Jan 19 at 0:00
AlarAlar
1
$endgroup$
How I can prove this inequality? How can I prove it using Cauchy–Schwarz inequality?
Let $a_1,...,a_n,b_1,....,b_n$ are any real numbers.
$sqrt {sum_{k=1}^n (a_k+b_k)^2)} leq sqrt {sum_{k=1}^n {a_k}^2} sqrt {sum_{k=1}^n {b_k}^2}$
real-analysis analysis inequality
real-analysis analysis inequality
edited Jan 19 at 0:09
Bernard
121k740116
asked Jan 19 at 0:00
AlarAlar
1
edited Jan 19 at 0:09
Bernard
121k740116
asked Jan 19 at 0:00
AlarAlar
1
edited Jan 19 at 0:09
Bernard
121k740116
edited Jan 19 at 0:09
Bernard
121k740116
edited Jan 19 at 0:09
Bernard
121k740116
121k740116
asked Jan 19 at 0:00
AlarAlar
1
asked Jan 19 at 0:00
AlarAlar
1
asked Jan 19 at 0:00
AlarAlar
1
1
add a comment |
add a comment |
1 Answer
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$begingroup$
This is false when $a_1 = b_1= 1$ and $a_k = b_k = 0$ for $k ne 1$.
answered Jan 19 at 0:05
angryavianangryavian
41.8k23381
$endgroup$
add a comment |
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$begingroup$
This is false when $a_1 = b_1= 1$ and $a_k = b_k = 0$ for $k ne 1$.
answered Jan 19 at 0:05
angryavianangryavian
41.8k23381
$endgroup$
add a comment |
$begingroup$
This is false when $a_1 = b_1= 1$ and $a_k = b_k = 0$ for $k ne 1$.
answered Jan 19 at 0:05
angryavianangryavian
41.8k23381
$endgroup$
add a comment |
$begingroup$
This is false when $a_1 = b_1= 1$ and $a_k = b_k = 0$ for $k ne 1$.
answered Jan 19 at 0:05
angryavianangryavian
41.8k23381
$endgroup$
This is false when $a_1 = b_1= 1$ and $a_k = b_k = 0$ for $k ne 1$.
answered Jan 19 at 0:05
angryavianangryavian
41.8k23381
answered Jan 19 at 0:05
angryavianangryavian
41.8k23381
answered Jan 19 at 0:05
angryavianangryavian
41.8k23381
answered Jan 19 at 0:05
angryavianangryavian
41.8k23381
41.8k23381
add a comment |
add a comment |
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