$a,b,cinBbb R^+, x,y,zin Bbb R, $ show that $fracx^2a+fracy^2b+fracz^2c ge frac(x+y+z)^2a+b+c$Prove $sumlimits_i=1^nfraca_i^2b_i geq frac(sumlimits_i=1^na_i)^2sumlimits_i=1^nb_i$Harmonic mean: show $maxax,by ge frac1a+b(x+y)$, $a,b>1$, $x,yge 0$Proof of analogue of the Cauchy-Schwarz inequality for random variablesOlympiad inequality $fraca2a + b + fracb2b + c + fracc2c + a leq 1$.Prove Inequality with Cauchy SchwarzShow that $frac1x_1(x_1+1)+frac1x_2(x_2+1)+cdots+frac1x_n(x_n+1)gefracn2$ with $x_1x_2cdots x_n=1$Estimating the product of timelike vectors in $Bbb R^n_nu$Showing that two vectors are not orthogonal, using Cauchy-Schwarz$AM$- $QM$ inequalityCauchy-Schwarz inequality to prove $A$ is spd$x,yinBbb R,$ find the maxima of $fracx+2y+3sqrtx^2+y^2+1$
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$a,b,cinBbb R^+, x,y,zin Bbb R, $ show that $fracx^2a+fracy^2b+fracz^2c ge frac(x+y+z)^2a+b+c$
Prove $sumlimits_i=1^nfraca_i^2b_i geq frac(sumlimits_i=1^na_i)^2sumlimits_i=1^nb_i$Harmonic mean: show $maxax,by ge frac1a+b(x+y)$, $a,b>1$, $x,yge 0$Proof of analogue of the Cauchy-Schwarz inequality for random variablesOlympiad inequality $fraca2a + b + fracb2b + c + fracc2c + a leq 1$.Prove Inequality with Cauchy SchwarzShow that $frac1x_1(x_1+1)+frac1x_2(x_2+1)+cdots+frac1x_n(x_n+1)gefracn2$ with $x_1x_2cdots x_n=1$Estimating the product of timelike vectors in $Bbb R^n_nu$Showing that two vectors are not orthogonal, using Cauchy-Schwarz$AM$- $QM$ inequalityCauchy-Schwarz inequality to prove $A$ is spd$x,yinBbb R,$ find the maxima of $fracx+2y+3sqrtx^2+y^2+1$
$begingroup$
$a,b,cinBbb R^+, x,y,zin Bbb R, $ show that $fracx^2a+fracy^2b+fracz^2c ge frac(x+y+z)^2a+b+c$ (use Cauchy–Schwarz inequality)
I have trouble finding the two vectors. Is it $(x,y,z)$ and $(a,b,c)$?
I need a hint
inequality cauchy-schwarz-inequality
edited 2 days ago
Michael Rozenberg
108k1895200
asked 2 days ago
DavidDavid
454
$endgroup$
add a comment |
$begingroup$
$a,b,cinBbb R^+, x,y,zin Bbb R, $ show that $fracx^2a+fracy^2b+fracz^2c ge frac(x+y+z)^2a+b+c$ (use Cauchy–Schwarz inequality)
I have trouble finding the two vectors. Is it $(x,y,z)$ and $(a,b,c)$?
I need a hint
inequality cauchy-schwarz-inequality
edited 2 days ago
Michael Rozenberg
108k1895200
asked 2 days ago
DavidDavid
454
$endgroup$
2
$begingroup$
This is known as Titus Lemma. More details here
$endgroup$
– rtybase
2 days ago
$begingroup$
Try $(x/sqrt a, y/sqrt b, z/sqrt c)$ and $(sqrt a, sqrt b, sqrt c)$…
$endgroup$
– Macavity
2 days ago
add a comment |
$begingroup$
$a,b,cinBbb R^+, x,y,zin Bbb R, $ show that $fracx^2a+fracy^2b+fracz^2c ge frac(x+y+z)^2a+b+c$ (use Cauchy–Schwarz inequality)
I have trouble finding the two vectors. Is it $(x,y,z)$ and $(a,b,c)$?
I need a hint
inequality cauchy-schwarz-inequality
edited 2 days ago
Michael Rozenberg
108k1895200
asked 2 days ago
DavidDavid
454
$endgroup$
$a,b,cinBbb R^+, x,y,zin Bbb R, $ show that $fracx^2a+fracy^2b+fracz^2c ge frac(x+y+z)^2a+b+c$ (use Cauchy–Schwarz inequality)
I have trouble finding the two vectors. Is it $(x,y,z)$ and $(a,b,c)$?
I need a hint
inequality cauchy-schwarz-inequality
inequality cauchy-schwarz-inequality
edited 2 days ago
Michael Rozenberg
108k1895200
asked 2 days ago
DavidDavid
454
edited 2 days ago
Michael Rozenberg
108k1895200
asked 2 days ago
DavidDavid
454
edited 2 days ago
Michael Rozenberg
108k1895200
edited 2 days ago
Michael Rozenberg
108k1895200
edited 2 days ago
Michael Rozenberg
108k1895200
108k1895200
asked 2 days ago
DavidDavid
454
asked 2 days ago
DavidDavid
454
asked 2 days ago
DavidDavid
454
454
2
$begingroup$
This is known as Titus Lemma. More details here
$endgroup$
– rtybase
2 days ago
$begingroup$
Try $(x/sqrt a, y/sqrt b, z/sqrt c)$ and $(sqrt a, sqrt b, sqrt c)$…
$endgroup$
– Macavity
2 days ago
add a comment |
2
$begingroup$
This is known as Titus Lemma. More details here
$endgroup$
– rtybase
2 days ago
$begingroup$
Try $(x/sqrt a, y/sqrt b, z/sqrt c)$ and $(sqrt a, sqrt b, sqrt c)$…
$endgroup$
– Macavity
2 days ago
2
2
$begingroup$
This is known as Titus Lemma. More details here
$endgroup$
– rtybase
2 days ago
$begingroup$
This is known as Titus Lemma. More details here
$endgroup$
– rtybase
2 days ago
$begingroup$
Try $(x/sqrt a, y/sqrt b, z/sqrt c)$ and $(sqrt a, sqrt b, sqrt c)$…
$endgroup$
– Macavity
2 days ago
$begingroup$
Try $(x/sqrt a, y/sqrt b, z/sqrt c)$ and $(sqrt a, sqrt b, sqrt c)$…
$endgroup$
– Macavity
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Use Cauchy Schwarz in Engelform:
$$fracx^2a+fracy^2b+fracz^2cgeq frac(x+y+z)^2a+b+c$$
It is equivalent to $$z^2a^2b+y^2a^2c+z^2ab^2-2,abcxy-2,abcxz-2,a
bcyz+y^2ac^2+x^2b^2c+x^2bc^2
geq 0$$
and this is $$a(bz-cy)^2+b(az-xc)^2+c(bx-ay)^2geq 0$$ if $$a,b,c$$ are positive.
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
add a comment |
$begingroup$
By C-S $$fracx^2a+fracy^2b+fracz^2c=fracleft(fracx^2a+fracy^2b+fracz^2cright)(a+b+c)a+b+cgeq$$
$$geqfracleft(fracxsqrtacdotsqrta+fracysqrtbcdotsqrtb+fraczsqrtccdotsqrtcright)^2a+b+c=frac(x+y+z)^2a+b+c.$$
The equality occurs for $$left(fracxsqrta,fracysqrtb,fraczsqrtcright)||(sqrta,sqrtb,sqrtc)$$ or
$$(x,y,z)||(a,b,c).$$
Actually, the last writing is very useful.
answered 2 days ago
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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active
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active
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votes
$begingroup$
Hint: Use Cauchy Schwarz in Engelform:
$$fracx^2a+fracy^2b+fracz^2cgeq frac(x+y+z)^2a+b+c$$
It is equivalent to $$z^2a^2b+y^2a^2c+z^2ab^2-2,abcxy-2,abcxz-2,a
bcyz+y^2ac^2+x^2b^2c+x^2bc^2
geq 0$$
and this is $$a(bz-cy)^2+b(az-xc)^2+c(bx-ay)^2geq 0$$ if $$a,b,c$$ are positive.
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
add a comment |
$begingroup$
Hint: Use Cauchy Schwarz in Engelform:
$$fracx^2a+fracy^2b+fracz^2cgeq frac(x+y+z)^2a+b+c$$
It is equivalent to $$z^2a^2b+y^2a^2c+z^2ab^2-2,abcxy-2,abcxz-2,a
bcyz+y^2ac^2+x^2b^2c+x^2bc^2
geq 0$$
and this is $$a(bz-cy)^2+b(az-xc)^2+c(bx-ay)^2geq 0$$ if $$a,b,c$$ are positive.
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
add a comment |
$begingroup$
Hint: Use Cauchy Schwarz in Engelform:
$$fracx^2a+fracy^2b+fracz^2cgeq frac(x+y+z)^2a+b+c$$
It is equivalent to $$z^2a^2b+y^2a^2c+z^2ab^2-2,abcxy-2,abcxz-2,a
bcyz+y^2ac^2+x^2b^2c+x^2bc^2
geq 0$$
and this is $$a(bz-cy)^2+b(az-xc)^2+c(bx-ay)^2geq 0$$ if $$a,b,c$$ are positive.
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
Hint: Use Cauchy Schwarz in Engelform:
$$fracx^2a+fracy^2b+fracz^2cgeq frac(x+y+z)^2a+b+c$$
It is equivalent to $$z^2a^2b+y^2a^2c+z^2ab^2-2,abcxy-2,abcxz-2,a
bcyz+y^2ac^2+x^2b^2c+x^2bc^2
geq 0$$
and this is $$a(bz-cy)^2+b(az-xc)^2+c(bx-ay)^2geq 0$$ if $$a,b,c$$ are positive.
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
edited 2 days ago
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
77.7k42866
add a comment |
add a comment |
$begingroup$
By C-S $$fracx^2a+fracy^2b+fracz^2c=fracleft(fracx^2a+fracy^2b+fracz^2cright)(a+b+c)a+b+cgeq$$
$$geqfracleft(fracxsqrtacdotsqrta+fracysqrtbcdotsqrtb+fraczsqrtccdotsqrtcright)^2a+b+c=frac(x+y+z)^2a+b+c.$$
The equality occurs for $$left(fracxsqrta,fracysqrtb,fraczsqrtcright)||(sqrta,sqrtb,sqrtc)$$ or
$$(x,y,z)||(a,b,c).$$
Actually, the last writing is very useful.
answered 2 days ago
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
By C-S $$fracx^2a+fracy^2b+fracz^2c=fracleft(fracx^2a+fracy^2b+fracz^2cright)(a+b+c)a+b+cgeq$$
$$geqfracleft(fracxsqrtacdotsqrta+fracysqrtbcdotsqrtb+fraczsqrtccdotsqrtcright)^2a+b+c=frac(x+y+z)^2a+b+c.$$
The equality occurs for $$left(fracxsqrta,fracysqrtb,fraczsqrtcright)||(sqrta,sqrtb,sqrtc)$$ or
$$(x,y,z)||(a,b,c).$$
Actually, the last writing is very useful.
answered 2 days ago
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
By C-S $$fracx^2a+fracy^2b+fracz^2c=fracleft(fracx^2a+fracy^2b+fracz^2cright)(a+b+c)a+b+cgeq$$
$$geqfracleft(fracxsqrtacdotsqrta+fracysqrtbcdotsqrtb+fraczsqrtccdotsqrtcright)^2a+b+c=frac(x+y+z)^2a+b+c.$$
The equality occurs for $$left(fracxsqrta,fracysqrtb,fraczsqrtcright)||(sqrta,sqrtb,sqrtc)$$ or
$$(x,y,z)||(a,b,c).$$
Actually, the last writing is very useful.
answered 2 days ago
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
By C-S $$fracx^2a+fracy^2b+fracz^2c=fracleft(fracx^2a+fracy^2b+fracz^2cright)(a+b+c)a+b+cgeq$$
$$geqfracleft(fracxsqrtacdotsqrta+fracysqrtbcdotsqrtb+fraczsqrtccdotsqrtcright)^2a+b+c=frac(x+y+z)^2a+b+c.$$
The equality occurs for $$left(fracxsqrta,fracysqrtb,fraczsqrtcright)||(sqrta,sqrtb,sqrtc)$$ or
$$(x,y,z)||(a,b,c).$$
Actually, the last writing is very useful.
answered 2 days ago
Michael RozenbergMichael Rozenberg
108k1895200
answered 2 days ago
Michael RozenbergMichael Rozenberg
108k1895200
answered 2 days ago
Michael RozenbergMichael Rozenberg
108k1895200
answered 2 days ago
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
add a comment |
add a comment |
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tk6qHveflNAMGr xgZQZCz0ngTbE7gJFirm n6P D,tDC9YQok,CrXPZ
2
$begingroup$
This is known as Titus Lemma. More details here
$endgroup$
– rtybase
2 days ago
$begingroup$
Try $(x/sqrt a, y/sqrt b, z/sqrt c)$ and $(sqrt a, sqrt b, sqrt c)$…
$endgroup$
– Macavity
2 days ago