$x,yinBbb R,$ find the maxima of $fracx+2y+3sqrtx^2+y^2+1$Inequality.$sqrtfrac2ab+c+sqrtfrac2bc+a+sqrtfrac2ca+b leq 3$How find this inequality $sqrta^2+64+sqrtb^2+1$Inequality. $fracasqrtb+fracbsqrtc+fraccsqrtageq3$How find this minimum $a+b+c+sqrta^2+b^2+sqrtb^2+c^2+sqrta^2+c^2$Prove that $fracab+2c+3d+fracbc+2d+3a+fraccd+2a+3b+fracda+2b+3c geq frac23.$Find the Maximum value of $fracxsqrtx+y+fracysqrty+z+fraczsqrtz+x$Find the minimize value of $A=frac1sqrta^2+b^2+c^2+1-frac2(a+1)(b+1)(c+1)$Find this maximum of the $fracsqrt34x^2+fracsqrt(9-x^2)(x^2-1)4$Cauchy–Schwarz inequality to find minimum of a functionFind minimum value of $sum frac sqrt asqrt b +sqrt c-sqrt a$
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$x,yinBbb R,$ find the maxima of $fracx+2y+3sqrtx^2+y^2+1$
Inequality.$sqrtfrac2ab+c}+sqrtfrac2bc+a+sqrt{frac2ca+b leq 3$How find this inequality $sqrta^2+64+sqrtb^2+1$Inequality. $fracasqrtb+fracbsqrtc+fraccsqrtageq3$How find this minimum $a+b+c+sqrta^2+b^2+sqrtb^2+c^2+sqrta^2+c^2$Prove that $fracab+2c+3d+fracbc+2d+3a+fraccd+2a+3b+fracda+2b+3c geq frac23.$Find the Maximum value of $fracxsqrtx+y+fracysqrty+z+fraczsqrtz+x$Find the minimize value of $A=frac1sqrta^2+b^2+c^2+1-frac2(a+1)(b+1)(c+1)$Find this maximum of the $fracsqrt34x^2+fracsqrt(9-x^2)(x^2-1)4$Cauchy–Schwarz inequality to find minimum of a functionFind minimum value of $sum frac sqrt asqrt b +sqrt c-sqrt a$
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I want to use Cauchy–Schwarz inequality, I sqared $fracx+2y+3sqrtx^2+y^2+1$ and got $fracx^2+4y^2+9x^2+y^2+1$, not sure if I am doing fine
real-analysis inequality maxima-minima cauchy-schwarz-inequality
edited 2 days ago
Michael Rozenberg
107k1895199
asked 2 days ago
DavidDavid
454
$endgroup$
add a comment |
$begingroup$
I want to use Cauchy–Schwarz inequality, I sqared $fracx+2y+3sqrtx^2+y^2+1$ and got $fracx^2+4y^2+9x^2+y^2+1$, not sure if I am doing fine
real-analysis inequality maxima-minima cauchy-schwarz-inequality
edited 2 days ago
Michael Rozenberg
107k1895199
asked 2 days ago
DavidDavid
454
$endgroup$
$begingroup$
Hint: One of the vectors should be $(x, y, 1)$.
$endgroup$
– PkT
2 days ago
2
$begingroup$
$x^2+4y^2+9$ is NOT the square of $x+2y+3$.
$endgroup$
– Crostul
2 days ago
$begingroup$
@Crostul Wow I didn't notice that serious mistake, thank you.
$endgroup$
– David
2 days ago
add a comment |
$begingroup$
I want to use Cauchy–Schwarz inequality, I sqared $fracx+2y+3sqrtx^2+y^2+1$ and got $fracx^2+4y^2+9x^2+y^2+1$, not sure if I am doing fine
real-analysis inequality maxima-minima cauchy-schwarz-inequality
edited 2 days ago
Michael Rozenberg
107k1895199
asked 2 days ago
DavidDavid
454
$endgroup$
I want to use Cauchy–Schwarz inequality, I sqared $fracx+2y+3sqrtx^2+y^2+1$ and got $fracx^2+4y^2+9x^2+y^2+1$, not sure if I am doing fine
real-analysis inequality maxima-minima cauchy-schwarz-inequality
real-analysis inequality maxima-minima cauchy-schwarz-inequality
edited 2 days ago
Michael Rozenberg
107k1895199
asked 2 days ago
DavidDavid
454
edited 2 days ago
Michael Rozenberg
107k1895199
asked 2 days ago
DavidDavid
454
edited 2 days ago
Michael Rozenberg
107k1895199
edited 2 days ago
Michael Rozenberg
107k1895199
edited 2 days ago
Michael Rozenberg
107k1895199
107k1895199
asked 2 days ago
DavidDavid
454
asked 2 days ago
DavidDavid
454
asked 2 days ago
DavidDavid
454
454
$begingroup$
Hint: One of the vectors should be $(x, y, 1)$.
$endgroup$
– PkT
2 days ago
2
$begingroup$
$x^2+4y^2+9$ is NOT the square of $x+2y+3$.
$endgroup$
– Crostul
2 days ago
$begingroup$
@Crostul Wow I didn't notice that serious mistake, thank you.
$endgroup$
– David
2 days ago
add a comment |
$begingroup$
Hint: One of the vectors should be $(x, y, 1)$.
$endgroup$
– PkT
2 days ago
2
$begingroup$
$x^2+4y^2+9$ is NOT the square of $x+2y+3$.
$endgroup$
– Crostul
2 days ago
$begingroup$
@Crostul Wow I didn't notice that serious mistake, thank you.
$endgroup$
– David
2 days ago
$begingroup$
Hint: One of the vectors should be $(x, y, 1)$.
$endgroup$
– PkT
2 days ago
$begingroup$
Hint: One of the vectors should be $(x, y, 1)$.
$endgroup$
– PkT
2 days ago
2
2
$begingroup$
$x^2+4y^2+9$ is NOT the square of $x+2y+3$.
$endgroup$
– Crostul
2 days ago
$begingroup$
$x^2+4y^2+9$ is NOT the square of $x+2y+3$.
$endgroup$
– Crostul
2 days ago
$begingroup$
@Crostul Wow I didn't notice that serious mistake, thank you.
$endgroup$
– David
2 days ago
$begingroup$
@Crostul Wow I didn't notice that serious mistake, thank you.
$endgroup$
– David
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Not that $$(x+2y+3)^2le (x^2+y^2+1)(1+4+9)$$
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
add a comment |
$begingroup$
By C-S
$$fracx+2y+3sqrtx^2+y^2+1leqsqrtfrac14(x+2y+3)^2(1^2+2^2+3^2)(x^2+y^2+1)leq$$
$$leqsqrtfrac14(x+2y+3)^2(x+2y+3)^2=sqrt14.$$
The equality occurs for $(x,y,1)||(1,2,3)$ and $x+2y+3>0,$ id est, for $(x,y)=left(frac13,frac23right),$
which says that we got a maximal value.
answered 2 days ago
Michael RozenbergMichael Rozenberg
107k1895199
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
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2 Answers
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$begingroup$
Not that $$(x+2y+3)^2le (x^2+y^2+1)(1+4+9)$$
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
add a comment |
$begingroup$
Not that $$(x+2y+3)^2le (x^2+y^2+1)(1+4+9)$$
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
add a comment |
$begingroup$
Not that $$(x+2y+3)^2le (x^2+y^2+1)(1+4+9)$$
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
Not that $$(x+2y+3)^2le (x^2+y^2+1)(1+4+9)$$
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
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
77.7k42866
add a comment |
add a comment |
$begingroup$
By C-S
$$fracx+2y+3sqrtx^2+y^2+1leqsqrtfrac14(x+2y+3)^2(1^2+2^2+3^2)(x^2+y^2+1)leq$$
$$leqsqrtfrac14(x+2y+3)^2(x+2y+3)^2=sqrt14.$$
The equality occurs for $(x,y,1)||(1,2,3)$ and $x+2y+3>0,$ id est, for $(x,y)=left(frac13,frac23right),$
which says that we got a maximal value.
answered 2 days ago
Michael RozenbergMichael Rozenberg
107k1895199
$endgroup$
add a comment |
$begingroup$
By C-S
$$fracx+2y+3sqrtx^2+y^2+1leqsqrtfrac14(x+2y+3)^2(1^2+2^2+3^2)(x^2+y^2+1)leq$$
$$leqsqrtfrac14(x+2y+3)^2(x+2y+3)^2=sqrt14.$$
The equality occurs for $(x,y,1)||(1,2,3)$ and $x+2y+3>0,$ id est, for $(x,y)=left(frac13,frac23right),$
which says that we got a maximal value.
answered 2 days ago
Michael RozenbergMichael Rozenberg
107k1895199
$endgroup$
add a comment |
$begingroup$
By C-S
$$fracx+2y+3sqrtx^2+y^2+1leqsqrtfrac14(x+2y+3)^2(1^2+2^2+3^2)(x^2+y^2+1)leq$$
$$leqsqrtfrac14(x+2y+3)^2(x+2y+3)^2=sqrt14.$$
The equality occurs for $(x,y,1)||(1,2,3)$ and $x+2y+3>0,$ id est, for $(x,y)=left(frac13,frac23right),$
which says that we got a maximal value.
answered 2 days ago
Michael RozenbergMichael Rozenberg
107k1895199
$endgroup$
By C-S
$$fracx+2y+3sqrtx^2+y^2+1leqsqrtfrac14(x+2y+3)^2(1^2+2^2+3^2)(x^2+y^2+1)leq$$
$$leqsqrtfrac14(x+2y+3)^2(x+2y+3)^2=sqrt14.$$
The equality occurs for $(x,y,1)||(1,2,3)$ and $x+2y+3>0,$ id est, for $(x,y)=left(frac13,frac23right),$
which says that we got a maximal value.
answered 2 days ago
Michael RozenbergMichael Rozenberg
107k1895199
answered 2 days ago
Michael RozenbergMichael Rozenberg
107k1895199
answered 2 days ago
Michael RozenbergMichael Rozenberg
107k1895199
answered 2 days ago
Michael RozenbergMichael Rozenberg
107k1895199
107k1895199
add a comment |
add a comment |
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$begingroup$
Hint: One of the vectors should be $(x, y, 1)$.
$endgroup$
– PkT
2 days ago
2
$begingroup$
$x^2+4y^2+9$ is NOT the square of $x+2y+3$.
$endgroup$
– Crostul
2 days ago
$begingroup$
@Crostul Wow I didn't notice that serious mistake, thank you.
$endgroup$
– David
2 days ago