Mixing
Inequality similar to Minkowski
$begingroup$
Prove that $(|x_1-z_1|^p+|x_2-z_2|^p)^{frac{1}{p}} le (|x_1-y_1|^p + |x_2-y_2|^p)^{frac{1}{p}}+(|y_1-z_1|^p + |y_2-z_2|^p)^{frac{1}{p}}$
The form is very close to Minkowski's inequality,but I can't really show this.
algebra-precalculus inequality
asked Jan 21 at 6:03
Legend KillerLegend Killer
1,6671524
$endgroup$
add a comment |
$begingroup$
Prove that $(|x_1-z_1|^p+|x_2-z_2|^p)^{frac{1}{p}} le (|x_1-y_1|^p + |x_2-y_2|^p)^{frac{1}{p}}+(|y_1-z_1|^p + |y_2-z_2|^p)^{frac{1}{p}}$
The form is very close to Minkowski's inequality,but I can't really show this.
algebra-precalculus inequality
asked Jan 21 at 6:03
Legend KillerLegend Killer
1,6671524
$endgroup$
add a comment |
$begingroup$
Prove that $(|x_1-z_1|^p+|x_2-z_2|^p)^{frac{1}{p}} le (|x_1-y_1|^p + |x_2-y_2|^p)^{frac{1}{p}}+(|y_1-z_1|^p + |y_2-z_2|^p)^{frac{1}{p}}$
The form is very close to Minkowski's inequality,but I can't really show this.
algebra-precalculus inequality
asked Jan 21 at 6:03
Legend KillerLegend Killer
1,6671524
$endgroup$
Prove that $(|x_1-z_1|^p+|x_2-z_2|^p)^{frac{1}{p}} le (|x_1-y_1|^p + |x_2-y_2|^p)^{frac{1}{p}}+(|y_1-z_1|^p + |y_2-z_2|^p)^{frac{1}{p}}$
The form is very close to Minkowski's inequality,but I can't really show this.
algebra-precalculus inequality
algebra-precalculus inequality
asked Jan 21 at 6:03
Legend KillerLegend Killer
1,6671524
asked Jan 21 at 6:03
Legend KillerLegend Killer
1,6671524
asked Jan 21 at 6:03
Legend KillerLegend Killer
1,6671524
asked Jan 21 at 6:03
Legend KillerLegend Killer
1,6671524
asked Jan 21 at 6:03
Legend KillerLegend Killer
1,6671524
1,6671524
add a comment |
add a comment |
1 Answer
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$begingroup$
First of all, you need $pgeq1$.
If so, it's just Minkowski because by the triangle inequality $$|x_1-y_1|+|y_1-z_1|geq|x_1-z_1|$$ and
$$|x_2-y_2|+|y_2-z_2|geq|x_2-z_2|.$$
answered Jan 21 at 6:15
Michael RozenbergMichael Rozenberg
107k1894198
$endgroup$
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1 Answer
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1 Answer
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$begingroup$
First of all, you need $pgeq1$.
If so, it's just Minkowski because by the triangle inequality $$|x_1-y_1|+|y_1-z_1|geq|x_1-z_1|$$ and
$$|x_2-y_2|+|y_2-z_2|geq|x_2-z_2|.$$
answered Jan 21 at 6:15
Michael RozenbergMichael Rozenberg
107k1894198
$endgroup$
add a comment |
$begingroup$
First of all, you need $pgeq1$.
If so, it's just Minkowski because by the triangle inequality $$|x_1-y_1|+|y_1-z_1|geq|x_1-z_1|$$ and
$$|x_2-y_2|+|y_2-z_2|geq|x_2-z_2|.$$
answered Jan 21 at 6:15
Michael RozenbergMichael Rozenberg
107k1894198
$endgroup$
add a comment |
$begingroup$
First of all, you need $pgeq1$.
If so, it's just Minkowski because by the triangle inequality $$|x_1-y_1|+|y_1-z_1|geq|x_1-z_1|$$ and
$$|x_2-y_2|+|y_2-z_2|geq|x_2-z_2|.$$
answered Jan 21 at 6:15
Michael RozenbergMichael Rozenberg
107k1894198
$endgroup$
First of all, you need $pgeq1$.
If so, it's just Minkowski because by the triangle inequality $$|x_1-y_1|+|y_1-z_1|geq|x_1-z_1|$$ and
$$|x_2-y_2|+|y_2-z_2|geq|x_2-z_2|.$$
answered Jan 21 at 6:15
Michael RozenbergMichael Rozenberg
107k1894198
edited Jan 21 at 9:53
answered Jan 21 at 6:15
Michael RozenbergMichael Rozenberg
107k1894198
answered Jan 21 at 6:15
Michael RozenbergMichael Rozenberg
107k1894198
answered Jan 21 at 6:15
Michael RozenbergMichael Rozenberg
107k1894198
107k1894198
add a comment |
add a comment |
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