Azim Premji University At Right Angles, November 2022 81
Student Corner
I
n this two-part article, we consider problems from various sources which are solved using the AM-GM inequality. We list the problems first and give the solutions later.Problems
Problem 1.Let a,b,c be positive numbers. Prove that (a
b )1/2
+ (b
c )1/3
+(c a
)1/5
>2. (1)
This problem is from Mathematical Reflection 2018; it was proposed by Florin Rotaru [1].
Problem 2.Let x,y,z,t be positive real numbers such that x+y+z+t=2. Show that
(4 x2 −1
)
· (4
y2 −1 )
· (4
z2 −1 )
· (4
t2 −1 )
≥ 154. (2) This problem was published in 2018 in Issue 1 of AMJ journal; it was proposed by Mihaela Berindeanu [2].
Problem 3.Let a,b,c,d be positive real numbers. Prove that a4+b4+c4+d4+4≥ 4
·(
(a2b2+1)(b2c2+1)(c2d2+1)(d2a2+1))1/4
. (3)
This problem was proposed by Angel Plaza and published in April 2014 in the SSMJ problem corner; it was solved by Albert Stadler [4].
1
Problems Based on the AM-GM Inequality – I
TOYESH PRAKASH SHARMA
Keywords: AM-GM inequality
82 Azim Premji University At Right Angles, November 2022
Solutions to problems 1, 2, 3
Problem 1.Let a,b,c be positive numbers. Prove that (a
b )1/2
+ (b
c )1/3
+(c a
)1/5
>2. Solution. From the AM-GM inequality we get
(a b
)1/2
+ (b
c )1/3
+(c a
)1/5
= (
2·1 2 ·a
b )1/2
+3·1 3 ·
(b c
)1/3
+5·1 5·(c
a )1/5
≥10· (1
22 · 1 33 · 1
55 ·a b ·b
c · c a
)1/10
=10· (1
22 · 1 33 · 1
55 )1/10
. So it suffices to prove that
1 22 · 1
33 · 1 55 > 1
510, or 55 >22·33, or 3125>108, which is clearly true.
Problem 2.Let x,y,z,t be positive real numbers such that x+y+z+t=2. Show that (4
x2 −1 )
· (4
y2 −1 )
· (4
z2 −1 )
· (4
t2 −1 )
≥154.
Solution. From the AM-GM inequality we get:
(x+y+z+t) +x≥5(
yztx2)1/5
, (x+y+z+t)−x≥3(yzt)1/3. Hence by multiplication,
(x+y+z+t)2−x2 ≥5(
yztx2)1/5
·3(yzt)1/3,
∴ 4−x2 ≥15·(
yztx2)1/5
·(yzt)1/3. In the same way we get:
4−y2≥ 15·(
xzty2)1/5
·(xzt)1/3, 4−z2≥ 15·(
yztz2)1/5
·(xyt)1/3, 4−t2≥ 15·(
yzxt2)1/5
·(xyz)1/3.
Azim Premji University At Right Angles, November 2022 83
Multiplication now yields:
(4−x2)
·( 4−y2)
·(
4−z2)
·( 4−t2)
≥154·x2·y2·z2·t2,
∴ ( 4 x2 −1
)
· (4
y2 −1 )
· (4
z2 −1 )
· (4
t2 −1 )
≥154. In [3], I published the following generalization of the above result:
Theorem(Toyesh Sharma). Let x1, x2, …, xnbe n positive real numbers, where n≥2, and let x1+x2+· · ·+xn=a; then the following inequality holds;
( a2 x21−1
)
· ( a2
x22−1 )
· · · ( a2
x2n−1 )
≥(
n2−1)n
. (4)
Problem 3.Let a,b,c,d be positive real numbers Prove that a4+b4+c4+d4+4≥4·(
(a2b2+1)(b2c2+1)(c2d2+1)(d2a2+1))1/4
. Solution. By the AM-GM inequality we get:
a4+b4≥2·a2·b2, b4+c4≥2·b2·c2, c4+d4≥2·c2·d2, d4+a4≥2·d2·a2. Adding these inequalities, we obtain
a4+b4+c4+d4+4≥(
a2b2+1) +(
b2c2+1) +(
c2d2+1) +(
d2a2+1) . By applying the AM-GM inequality once more, we obtain
(a2b2+1) +(
b2c2+1) +(
c2d2+1) +(
d2a2+1)
≥4((
a2b2+1)
·(
b2c2+1)
·(
c2d2+1)
·(
d2a2+1))1/4
, and the claim follows.
References
1. Florin Rotaru, Mathematical Reflections 2009, J468-Solution, Issue-1,
https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2019-01/mr_6_2018_solutions_2.pdf 2. Mihaela Berindeanu, “Problem EM-55”, Arhimede math., j. 5.1 (2018), pg. 33
3. Toyesh Prakash Sharma, “Generalization of Problem E 55”, Arhimede math., 7.2 (2018), pg. 136 4. Albert Stadler, “Prob. 5303, Angel Plaza”, SSMJ problem solution corner, Nov. 2014, pg. 7-9
TOYESH PRAKASH SHARMA has been interested in science, mathematics and literature since high school.
He passed out from St. C.F. Andrews School, Agra. In standard 11, he began doing research in the field of mathematics and has contributed articles to different journals and magazines such as Mathematical Gazette, Crux Mathematicorum, Parabola, AMJ, ISROSET, SSMJ, Pentagon, Octagon, La Gaceta de la RSME, At Right Angle, Fibonacci Quarterly, Mathematical Reflections etc. Currently he is doing his under graduation in the field of Physics and Mathematics from Agra College, Agra, India. He may be contacted at [email protected]
