387. Algebra (II).
by Virgil Nicula, Sep 27, 2013, 6:10 AM
PP13. Prove that 
.
Proof. Denote
and 
. Therefore:
.
Hence
. In conclusion, 
.
PP14. Find all pairs (x,y) of real numbers such that
Proof.
![$2\cdot \left[16^{\left(x^2+y\right)+\left(y^2+x\right)}\right]^{\frac{1}{2}} \ge $](//latex.artofproblemsolving.com/1/6/9/16987f042a3e3d7ebffdad8cf6b4d0a6e21e9362.png)
. This gives the case of equality so that 
. So, 
which is true when put into the original equation.
PP15. Find real numbers
so that 
.
Proof. Denote the set
of the zeroes of 
, where ![$f(x)=2^x+2^{\sqrt{1-x^2}}\ ,\ x\in[0,1]$](//latex.artofproblemsolving.com/e/6/1/e618bcda6d976dd17cf752e44f676e49082706b4.png)
. Thus, 
, i.e. ![$\{0,1\}\subset \mathbb S\subset[0,1]$](//latex.artofproblemsolving.com/e/d/5/ed53fcbcce2ef4cb6d63d865b717cde9f9f8454e.png)
. Let 
. Thus,
![$\left\{\begin{array}{cccccccc}
f(x)=2\cdot 2^{x-1}+2^{\sqrt{1-x^2}} & \implies & 3\ge 3\cdot\sqrt[3]{2^{2(x-1)+\sqrt{1-x^2}}} & \implies & 2(x-1)+\sqrt{1-x^2}\le 0 & \implies & x\le \frac 35\\\\
f(x)=2^x+2\cdot 2^{\sqrt{1-x^2}-1} & \implies & 3\ge 3\cdot\sqrt[3]{2^{x+2\left(\sqrt{1-x^2}-1\right)}} & \implies & x+2\left(\sqrt{1-x^2}-1\right)\le 0 & \implies & x\ge\frac 45\end{array}\right\|$](//latex.artofproblemsolving.com/7/4/9/749a8bd1cb328e0e176bd9cea912f0623631b7a3.png)
![$\implies x\in \left(0,\frac 35\right]\cap\left[\frac 45,1\right)=\emptyset$](//latex.artofproblemsolving.com/f/d/2/fd24a63b07edef12b7d60f560c98e3a63f1a22ba.png)
, i.e. 
.
PP16. Let
. Prove that 
.
Proof 1.![$f(x)=(x-a)g(x)\ ,\ g(x)=\left[x^2+ax+\left(a^2-21\right)\right]$](//latex.artofproblemsolving.com/e/f/4/ef47b7cf428a7f741c0f04c44a52b8e6aee229a0.png)
. Thus, 
, O.K.
Proof 2. Observe that
and 
. Denote 
. Therefore, 
and
![$\left(a^2-21\right)+\left[\left(a^2-14\right)^2+5a\left(a^2-14\right)+6a^2\right]=$](//latex.artofproblemsolving.com/1/e/d/1ed4ccf260df89a12aba56a9323acf8af5604897.png)
. In conclusion, 
, i.e. 
.
Proof 3. Denote
where 
and 
. Thus, 
have at least one common root 
.
Proof 4. Observe that and . Thus,
![$\left[\left(a^2-14\right)^2+5a\left(a^2-14\right)+6a^2\right]+\left(a^2-21\right)=0\iff$](//latex.artofproblemsolving.com/1/4/b/14b23e4d75964f56387dc379fe416b84a7b9e7d4.png)
, what is truly.
Proof 5. Observe that and . Thus, 
and for
obtain that 
. In conclusion,
, what is truly because 
and 
.
Remark.
what is truly.
PP17. Let
such that 
. Prove that that 
.
Proof 1. If all of
are non-negative, then the inequality is obvious. Suppose w.l.o.g. that 
. Then 
. Write the inequality as 
. This is a convex quadratic in 
because 
. The inequality is true 
. But 
because
, so we are done.
Proof 2.![$0\le \left[(bz+cy)-a(y+z)\right]^2+(b-c)^2(xy+yz+zx)=$](//latex.artofproblemsolving.com/c/f/e/cfeb099f65c7aa77b04e42fb7a6e7dc7c4b78801.png)
![$\left[y(a-c)+z(a-b)\right]^2+(b-c)^2(xy+yz+zx)=$](//latex.artofproblemsolving.com/8/8/a/88a14d8499a944e1accc0433005b35aebf2055fa.png)
![$(y+z)\left[a^2(y+z)+b^2(z+x)+c^2(y+x)-2bcx-2acy-2abz\right]=(y+z)\left[x(b-c)^2 + y(c-a)^2 + z(a-b)^2 \right]$](//latex.artofproblemsolving.com/5/b/2/5b254085718e83918839ff3ef5af496847a3e204.png)
and 
a.s.o.
PP18. Prove that for any real there is the equivalence 
![$ (1+x)^2\ \iff\ x\in\left[-\frac 52,-1\right]\cup\left\{\sqrt 2-1\right\}$](//latex.artofproblemsolving.com/d/8/c/d8ce186dc9b86e7e9b607b7e6175abe59e5b11fb.png)
.
Proof. Observe that![$x\in\left[-\frac 52,\frac 12\right]$](//latex.artofproblemsolving.com/5/1/c/51c64ab53ea328af18b591edf718750a4debf7c5.png)
, i.e. 
. Denote 
, where 
. Our inequation becomes 
, where . Since
, then for any ![$y\in\left[-\frac 32,0\right]$](//latex.artofproblemsolving.com/2/4/3/243c7e035e4c6dec777943a2143035fea7aed8f2.png)
the inequality 
is truly. Suppose that ![$y\in \left(0,\frac 32\right]$](//latex.artofproblemsolving.com/3/9/3/393fadbe4e3ee7e79e5d88437b1feafffc5cc32d.png)
. In this case, 
and the inequality is equivalently with 
and 
. In conclusion, ![$y\in\left[-\frac 32,0\right]\cup \left\{\sqrt 2\right\}$](//latex.artofproblemsolving.com/3/6/4/36449fe2770f6243d4e5404ab7248e391e250a2d.png)
, i.e.
![$x\in\left[-\frac 52,-1\right]\cup\left\{\sqrt 2-1\right\}$](//latex.artofproblemsolving.com/c/1/2/c12da29d5fd6ddf75fa9acd52ad20daec1770290.png)
.
PP19. Let
so that 
. Prove that 
.
Proof 1. Denote
. Thus, 
. Suppose that 
.
Thus,
, what is falsely. In conclusion, 
, i.e. .
Proof 2.
, where 
. Thus, 
![$m\left[p^2+2(n-q)\right]-3mn=p^3-3pq\implies$](//latex.artofproblemsolving.com/7/2/2/72293d3e8b6e5a6535b71b6e1f88c2d278900edf.png)
![$\left[\frac {p\left(p^2-3q\right)}{p^2-(n+2q)}\right]^2=p^2+2(n-q)\implies$](//latex.artofproblemsolving.com/1/a/b/1ab02788074a7fadacfb8d8a4467884243db96d7.png)
![$p^2\left(p^2-3q\right)^2=\left[p^2+2(n-q)\right]\cdot\left[p^2-(n+2q)\right]^2\implies$](//latex.artofproblemsolving.com/7/4/a/74a99f31d38201376e268d01205f7fddeb95ffef.png)
. Suppose w.l.o.g. that 
. Thus,
, what is falsely. In conclusion, 
, i.e. .
Proof 3.
and 
so that 
. Observe that 
and
. Denote the function 
. Prove easily that
we have 
, i.e. the line 
is symmetry axis for the graph 
and on 
the function 
is decreasing 
.
In conclusion,
, where 
.
PP20. Let
with the roots 
. Denote 
, where 
. Determine the product 
.
Proof 1 (mavropnevma). Let
. Then 
. So 
.
But
. So 
.
Proof 2. I"ll ascertain the equation with the roots
. Eliminate the variable between the relations 
and 
. Therefore, 
and
and 
. Hence 
![$\left[-\frac {y+3}{(y+2)^2}\right]^2=y+2\implies$](//latex.artofproblemsolving.com/5/9/4/5945de4ee962d446183eba8ae525a8c4b1f27bb5.png)
the equation 
has the roots . In conclusion, 
.
PP21. Let with the roots . Find the equation with the roots 
, where and .
Proof 1 (mavropnevma). Let . Take 
a primitive complex 
-root of unity. Thus 
and 
.
Then
![$ \left[\prod_{k=1}^5\left(\omega-x_k\right)\right] \left [\prod_{k=1}^5\left(\overline{\omega}-x_k\right)\right] =$](//latex.artofproblemsolving.com/e/9/7/e9736d6c3a6a0ce9ab3c2e685acd1cdf1417464d.png)
. But 
.
Proof 2. Eliminate between the relations 
and obtain the following chain: 
.
Now see PP12 from here. The equations
have at least one common root
![$\left[2(y-1)^2-1\right]^2+(1-y)(2+y)\left[y(1-y)^2+1\right]=0\implies$](//latex.artofproblemsolving.com/7/9/f/79f15c1ac83d05e36197d471b0fa15c719a40147.png)
, what is the equation with the roots 
and 
.
PP22. Let
, i.e. 
. Evaluate 
.
Proof 1. Denote
. Observe that 
and 
and 
. Thus, ![$p=\prod_{k=1}^3y_k=\prod_{k=1}^3\left[(-3)\cdot\frac {-\frac 23-x_k}{1-x_k}\right]\implies$](//latex.artofproblemsolving.com/3/9/5/3955b78d5b34057ed53c0841278abb434d22c28c.png)
.
Proof 2. Let
. Take a primitive complex -root of unity. Thus and . Then 
![$ \left[\prod_{k=1}^3\left(\omega-x_k\right)\right] \left [\prod_{k=1}^3\left(\overline{\omega}-x_k\right)\right] =$](//latex.artofproblemsolving.com/7/8/0/78083d4b5c4f797ab4f15cad8a91683b4934710a.png)
. But 
Proof 3 (general). Eliminate between 
. I"ll obtain the equation with the roots 
. Thus, equations have at least a common root 
.
An easy extension. Consider the equation
, where 
. Denote 
. Prove that 
and 
(constant).
Remark. Can prove that the equation with the roots
is 
.
PP23. Let
for which 
. Prove that ![$\{x,y,z\}\subset \left[-\frac 13,5\right]$](//latex.artofproblemsolving.com/f/a/0/fa0ecccfd8695d4528346b26079c60753eff23ce.png)
.
Proof 1.
. Since 
obtain that 
.
Proof 2. Denote
. Therefore, 
. Thus,
![$z\in \left[-\frac 13,5\right]$](//latex.artofproblemsolving.com/b/5/5/b556995c57c7ce7d86f3511ccede2592852a3561.png)
and by symmetry obtain that .
Proof 3. Denote the equation
with 
. Observe that 
and
, i.e. 
. Since 
and 
obtain that
because 
PP24. Find
if the numbers 
satisfy the equations 
Proof 1. Set
and denote 
. Thus, 
I"ll use the identity
, i.e. 
. Therefore,
In conclusion,
.
Proof 2. The homogeneous system
with the variables 
is compatibly, where 
a.s.o.
.Proof. Denote
and 
. Therefore:
.
Hence
. In conclusion, 
.PP14. Find all pairs (x,y) of real numbers such that
Proof.
![$2\cdot \left[16^{\left(x^2+y\right)+\left(y^2+x\right)}\right]^{\frac{1}{2}} \ge $](http://latex.artofproblemsolving.com/1/6/9/16987f042a3e3d7ebffdad8cf6b4d0a6e21e9362.png)
. This gives the case of equality so that 
. So, 
which is true when put into the original equation. PP15. Find real numbers
so that 
.Proof. Denote the set
of the zeroes of 
, where ![$f(x)=2^x+2^{\sqrt{1-x^2}}\ ,\ x\in[0,1]$](http://latex.artofproblemsolving.com/e/6/1/e618bcda6d976dd17cf752e44f676e49082706b4.png)
. Thus, 
, i.e. ![$\{0,1\}\subset \mathbb S\subset[0,1]$](http://latex.artofproblemsolving.com/e/d/5/ed53fcbcce2ef4cb6d63d865b717cde9f9f8454e.png)
. Let 
. Thus,![$\left\{\begin{array}{cccccccc}
f(x)=2\cdot 2^{x-1}+2^{\sqrt{1-x^2}} & \implies & 3\ge 3\cdot\sqrt[3]{2^{2(x-1)+\sqrt{1-x^2}}} & \implies & 2(x-1)+\sqrt{1-x^2}\le 0 & \implies & x\le \frac 35\\\\
f(x)=2^x+2\cdot 2^{\sqrt{1-x^2}-1} & \implies & 3\ge 3\cdot\sqrt[3]{2^{x+2\left(\sqrt{1-x^2}-1\right)}} & \implies & x+2\left(\sqrt{1-x^2}-1\right)\le 0 & \implies & x\ge\frac 45\end{array}\right\|$](http://latex.artofproblemsolving.com/7/4/9/749a8bd1cb328e0e176bd9cea912f0623631b7a3.png)
![$\implies x\in \left(0,\frac 35\right]\cap\left[\frac 45,1\right)=\emptyset$](http://latex.artofproblemsolving.com/f/d/2/fd24a63b07edef12b7d60f560c98e3a63f1a22ba.png)
, i.e. 
.PP16. Let
. Prove that 
. Proof 1.
![$f(x)=(x-a)g(x)\ ,\ g(x)=\left[x^2+ax+\left(a^2-21\right)\right]$](http://latex.artofproblemsolving.com/e/f/4/ef47b7cf428a7f741c0f04c44a52b8e6aee229a0.png)
. Thus, 
, O.K.Proof 2. Observe that
and 
. Denote 
. Therefore, 
and
![$\left(a^2-21\right)+\left[\left(a^2-14\right)^2+5a\left(a^2-14\right)+6a^2\right]=$](http://latex.artofproblemsolving.com/1/e/d/1ed4ccf260df89a12aba56a9323acf8af5604897.png)
. In conclusion, 
, i.e. 
.Proof 3. Denote
where 
and 
. Thus, 
have at least one common root 
.Proof 4. Observe that
and . Thus, 
![$\left[\left(a^2-14\right)^2+5a\left(a^2-14\right)+6a^2\right]+\left(a^2-21\right)=0\iff$](http://latex.artofproblemsolving.com/1/4/b/14b23e4d75964f56387dc379fe416b84a7b9e7d4.png)
, what is truly.Proof 5. Observe that
and . Thus, 
and for
obtain that 
. In conclusion, 
, what is truly because 
and 
.Remark.
what is truly.PP17. Let
such that 
. Prove that that 
. Proof 1. If all of
are non-negative, then the inequality is obvious. Suppose w.l.o.g. that 
. Then 
. Write the inequality as 
. This is a convex quadratic in 
because 
. The inequality is true 
. But 
because
, so we are done.Proof 2.
![$0\le \left[(bz+cy)-a(y+z)\right]^2+(b-c)^2(xy+yz+zx)=$](http://latex.artofproblemsolving.com/c/f/e/cfeb099f65c7aa77b04e42fb7a6e7dc7c4b78801.png)
![$\left[y(a-c)+z(a-b)\right]^2+(b-c)^2(xy+yz+zx)=$](http://latex.artofproblemsolving.com/8/8/a/88a14d8499a944e1accc0433005b35aebf2055fa.png)
![$(y+z)\left[a^2(y+z)+b^2(z+x)+c^2(y+x)-2bcx-2acy-2abz\right]=(y+z)\left[x(b-c)^2 + y(c-a)^2 + z(a-b)^2 \right]$](http://latex.artofproblemsolving.com/5/b/2/5b254085718e83918839ff3ef5af496847a3e204.png)
and 
a.s.o.PP18. Prove that for any real
there is the equivalence 
![$ (1+x)^2\ \iff\ x\in\left[-\frac 52,-1\right]\cup\left\{\sqrt 2-1\right\}$](http://latex.artofproblemsolving.com/d/8/c/d8ce186dc9b86e7e9b607b7e6175abe59e5b11fb.png)
.Proof. Observe that
![$x\in\left[-\frac 52,\frac 12\right]$](http://latex.artofproblemsolving.com/5/1/c/51c64ab53ea328af18b591edf718750a4debf7c5.png)
, i.e. 
. Denote 
, where 
. Our inequation becomes 
, where . Since
, then for any ![$y\in\left[-\frac 32,0\right]$](http://latex.artofproblemsolving.com/2/4/3/243c7e035e4c6dec777943a2143035fea7aed8f2.png)
the inequality 
is truly. Suppose that ![$y\in \left(0,\frac 32\right]$](http://latex.artofproblemsolving.com/3/9/3/393fadbe4e3ee7e79e5d88437b1feafffc5cc32d.png)
. In this case, 
and the inequality
is equivalently with 
and 
. In conclusion, ![$y\in\left[-\frac 32,0\right]\cup \left\{\sqrt 2\right\}$](http://latex.artofproblemsolving.com/3/6/4/36449fe2770f6243d4e5404ab7248e391e250a2d.png)
, i.e.![$x\in\left[-\frac 52,-1\right]\cup\left\{\sqrt 2-1\right\}$](http://latex.artofproblemsolving.com/c/1/2/c12da29d5fd6ddf75fa9acd52ad20daec1770290.png)
. PP19. Let
so that 
. Prove that 
.Proof 1. Denote
. Thus, 
. Suppose that 
.Thus,
, what is falsely. In conclusion, 
, i.e. .Proof 2.
, where 
. Thus, 
![$m\left[p^2+2(n-q)\right]-3mn=p^3-3pq\implies$](http://latex.artofproblemsolving.com/7/2/2/72293d3e8b6e5a6535b71b6e1f88c2d278900edf.png)
![$\left[\frac {p\left(p^2-3q\right)}{p^2-(n+2q)}\right]^2=p^2+2(n-q)\implies$](http://latex.artofproblemsolving.com/1/a/b/1ab02788074a7fadacfb8d8a4467884243db96d7.png)
![$p^2\left(p^2-3q\right)^2=\left[p^2+2(n-q)\right]\cdot\left[p^2-(n+2q)\right]^2\implies$](http://latex.artofproblemsolving.com/7/4/a/74a99f31d38201376e268d01205f7fddeb95ffef.png)
. Suppose w.l.o.g. that 
. Thus,
, what is falsely. In conclusion, 
, i.e. .Proof 3.
and 
so that 
. Observe that 
and
. Denote the function 
. Prove easily that
we have 
, i.e. the line 
is symmetry axis for the graph 
and on 
the function 
is decreasing 
.In conclusion,
, where 
. PP20. Let
with the roots 
. Denote 
, where 
. Determine the product 
.Proof 1 (mavropnevma). Let
. Then 
. So 
.But
. So 
.Proof 2. I"ll ascertain the equation with the roots
. Eliminate the variable between the relations 
and 
. Therefore, 
and
and 
. Hence 
![$\left[-\frac {y+3}{(y+2)^2}\right]^2=y+2\implies$](http://latex.artofproblemsolving.com/5/9/4/5945de4ee962d446183eba8ae525a8c4b1f27bb5.png)
the equation 
has the roots . In conclusion, 
.PP21. Let
with the roots . Find the equation with the roots 
, where and .Proof 1 (mavropnevma). Let
. Take 
a primitive complex 
-root of unity. Thus 
and 
.Then
![$ \left[\prod_{k=1}^5\left(\omega-x_k\right)\right] \left [\prod_{k=1}^5\left(\overline{\omega}-x_k\right)\right] =$](http://latex.artofproblemsolving.com/e/9/7/e9736d6c3a6a0ce9ab3c2e685acd1cdf1417464d.png)
. But 
.Proof 2. Eliminate
between the relations 
and obtain the following chain: 
.Now see PP12 from here. The equations
have at least one common root 
![$\left[2(y-1)^2-1\right]^2+(1-y)(2+y)\left[y(1-y)^2+1\right]=0\implies$](http://latex.artofproblemsolving.com/7/9/f/79f15c1ac83d05e36197d471b0fa15c719a40147.png)
, what is the equation with the roots 
and 
.PP22. Let
, i.e. 
. Evaluate 
.Proof 1. Denote
. Observe that 
and 
and 
. Thus, ![$p=\prod_{k=1}^3y_k=\prod_{k=1}^3\left[(-3)\cdot\frac {-\frac 23-x_k}{1-x_k}\right]\implies$](http://latex.artofproblemsolving.com/3/9/5/3955b78d5b34057ed53c0841278abb434d22c28c.png)
.Proof 2. Let
. Take a primitive complex -root of unity. Thus and . Then 
![$ \left[\prod_{k=1}^3\left(\omega-x_k\right)\right] \left [\prod_{k=1}^3\left(\overline{\omega}-x_k\right)\right] =$](http://latex.artofproblemsolving.com/7/8/0/78083d4b5c4f797ab4f15cad8a91683b4934710a.png)
. But 
Proof 3 (general). Eliminate
between 
. I"ll obtain the equation with the roots 
. Thus, equations have at least a common root 
.An easy extension. Consider the equation
, where 
. Denote 
. Prove that 
and 
(constant).Remark. Can prove that the equation with the roots
is 
.PP23. Let
for which 
. Prove that ![$\{x,y,z\}\subset \left[-\frac 13,5\right]$](http://latex.artofproblemsolving.com/f/a/0/fa0ecccfd8695d4528346b26079c60753eff23ce.png)
.Proof 1.
. Since 
obtain that 
.Proof 2. Denote
. Therefore, 
. Thus,
![$z\in \left[-\frac 13,5\right]$](http://latex.artofproblemsolving.com/b/5/5/b556995c57c7ce7d86f3511ccede2592852a3561.png)
and by symmetry obtain that .Proof 3. Denote the equation
with 
. Observe that 
and
, i.e. 
. Since 
and 
obtain that because 
PP24. Find
if the numbers 
satisfy the equations 
Proof 1. Set
and denote 
. Thus, 
I"ll use the identity
, i.e. 
. Therefore,
In conclusion,
.Proof 2. The homogeneous system
with the variables 
is compatibly, where 
a.s.o.This post has been edited 229 times. Last edited by Virgil Nicula, Feb 11, 2018, 3:10 PM
August 2018
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