Prove that for positive reals a , b , c a,b,c a , b , c summing up to 1 1 1 , we have
1 a + b + 16 c + 81 a + b + c ≥ 98. \dfrac{1}{a+b}+\dfrac{16}{c}+\dfrac{81}{a+b+c}\ge 98. a + b 1 + c 1 6 + a + b + c 8 1 ≥ 9 8 .
The hypothesis implies a + b + c = 1 a+b+c=1 a + b + c = 1 . We have square terms in the numerator, so
1 a + b + 16 c + 81 a + b + c = 1 a + b + 4 2 c + 9 2 a + b + c ≥ ( 1 + 4 + 9 ) 2 ( a + b ) + c + ( a + b + c ) = ( 14 ) 2 2 ( a + b + c ) = 196 2 = 98 , \begin{aligned}
\dfrac{1}{a+b}+\dfrac{16}{c}+\dfrac{81}{a+b+c} &=& \dfrac{1}{a+b}+\dfrac{4^2}{c}+\dfrac{9^2}{a+b+c} \\ \\
&\ge &\dfrac{(1+4+9)^2}{(a+b)+c+(a+b+c)} \\ \\
& = & \dfrac{(14)^2}{2(a+b+c)} = \dfrac{196}{2} = 98,
\end{aligned} a + b 1 + c 1 6 + a + b + c 8 1 = ≥ = a + b 1 + c 4 2 + a + b + c 9 2 ( a + b ) + c + ( a + b + c ) ( 1 + 4 + 9 ) 2 2 ( a + b + c ) ( 1 4 ) 2 = 2 1 9 6 = 9 8 ,
which is exactly what we wanted. □ _\square □
Note that the equality condition a + b = k , c = 16 k , a + b + c = 81 k a+b = k, c = 16k, a+b+c = 81k a + b = k , c = 1 6 k , a + b + c = 8 1 k cannot be satisfied.
In fact, we have the stronger statement:
1 a + b + 16 c + 81 a + b + c ≥ 106. \dfrac{1}{a+b}+\dfrac{16}{c}+\dfrac{81}{a+b+c}\ge 106. a + b 1 + c 1 6 + a + b + c 8 1 ≥ 1 0 6 .
Can you prove this using Titu's lemma?
Over all triplets of positive reals x , y , z x,y,z x , y , z satisfying x + y + z = 3 x+y+z=3 x + y + z = 3 , find the minimum value of
f ( x , y , z ) = y z + 4 z x + 9 x y x y z . f(x,y,z)=\dfrac{yz+4zx+9xy}{xyz}. f ( x , y , z ) = x y z y z + 4 z x + 9 x y .
Dividing out simplifies f ( x , y , z ) f(x,y,z) f ( x , y , z ) into 1 x + 4 y + 9 z \frac{1}{x}+\frac{4}{y}+\frac{9}{z} x 1 + y 4 + z 9 . Now we have square terms on the numerator. We get
f ( x , y , z ) = 1 2 x + 2 2 y + 3 2 z ≥ ( 1 + 2 + 3 ) 2 x + y + z = 36 3 = 12. f(x,y,z)=\dfrac{1^2}{x}+\dfrac{2^2}{y}+\dfrac{3^2}{z} \ge \dfrac{(1+2+3)^2}{x+y+z} = \dfrac{36}{3}=12. f ( x , y , z ) = x 1 2 + y 2 2 + z 3 2 ≥ x + y + z ( 1 + 2 + 3 ) 2 = 3 3 6 = 1 2 .
Equality holds when
1 x = 2 y = 3 z ⟺ y = 2 x , z = 3 y 2 = 3 x , \dfrac 1 x = \dfrac 2 y = \dfrac 3 z~\Longleftrightarrow ~ y=2x, ~z=\dfrac{3y}{2}=3x, x 1 = y 2 = z 3 ⟺ y = 2 x , z = 2 3 y = 3 x ,
and solving for x x x using the constraint x + y + z = 3 x+y+z=3 x + y + z = 3 , we have equality at x = 1 2 x=\frac{1}{2} x = 2 1 , y = 1 y=1 y = 1 , z = 3 2 z=\frac{3}{2} z = 2 3 . We check that equality indeed occurs.
Therefore, the minimum value is 12. □ _\square □
Let x , y , z ∈ R + x,y,z\in\mathbb{R^{+}} x , y , z ∈ R + . Prove that
2 x + y + 2 y + z + 2 z + x ≥ 9 x + y + z . \frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}≥\frac{9}{x+y+z}. x + y 2 + y + z 2 + z + x 2 ≥ x + y + z 9 .
Observe that
2 x + y + 2 y + z + 2 z + x = ( 2 ) 2 x + y + ( 2 ) 2 y + z + ( 2 ) 2 z + x . \frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}=\frac{\big(\sqrt{2}\big)^{2}}{x+y}+\frac{\big(\sqrt{2}\big)^{2}}{y+z}+\frac{\big(\sqrt{2}\big)^{2}}{z+x}. x + y 2 + y + z 2 + z + x 2 = x + y ( 2 ) 2 + y + z ( 2 ) 2 + z + x ( 2 ) 2 .
Then,
( 2 ) 2 x + y + ( 2 ) 2 y + z + ( 2 ) 2 z + x ≥ ( 3 2 ) 2 2 ( x + y + z ) = 9 x + y + z . □ \frac{\big(\sqrt{2}\big)^{2}}{x+y}+\frac{\big(\sqrt{2}\big)^{2}}{y+z}+\frac{\big(\sqrt{2}\big)^{2}}{z+x}≥\frac{\big(3\sqrt{2}\big)^{2}}{2(x+y+z)}=\frac{9}{x+y+z}. \ _\square x + y ( 2 ) 2 + y + z ( 2 ) 2 + z + x ( 2 ) 2 ≥ 2 ( x + y + z ) ( 3 2 ) 2 = x + y + z 9 . □
Let a , b ∈ R + a,b\in\mathbb{R^{+}} a , b ∈ R + . Prove that
8 ( a 4 + b 4 ) ≥ ( a + b ) 4 . 8\big(a^{4}+b^{4}\big)\geq(a+b)^{4}. 8 ( a 4 + b 4 ) ≥ ( a + b ) 4 .
We have
a 4 + b 4 = a 4 1 + b 4 1 ≥ ( a 2 + b 2 ) 2 2 ≥ ( ( a + b ) 2 2 ) 2 2 = ( a + b ) 4 8 . □ a^{4}+b^{4}=\dfrac{a^{4}}{1}+\dfrac{b^{4}}{1}\geq\dfrac{\left(a^{2}+b^{2}\right)^{2}}{2}\geq\dfrac{\Big(\frac{(a+b)^{2}}{2}\Big)^{2}}{2}=\dfrac{(a+b)^{4}}{8}. \ _\square a 4 + b 4 = 1 a 4 + 1 b 4 ≥ 2 ( a 2 + b 2 ) 2 ≥ 2 ( 2 ( a + b ) 2 ) 2 = 8 ( a + b ) 4 . □
For a , b , c ∈ R + a,b,c\in\mathbb{R^{+}} a , b , c ∈ R + , prove that
a 2 + b 2 a + b + b 2 + c 2 b + c + c 2 + a 2 c + a ≥ a + b + c . \frac{a^{2}+b^{2}}{a+b}+\frac{b^{2}+c^{2}}{b+c}+\frac{c^{2}+a^{2}}{c+a}≥a+b+c. a + b a 2 + b 2 + b + c b 2 + c 2 + c + a c 2 + a 2 ≥ a + b + c .
Write the left side of the given inequality as
a 2 a + b + b 2 a + b + b 2 b + c + a 2 c + a + c 2 a + c + c 2 c + b , \dfrac{a^{2}}{a+b}+\dfrac{b^{2}}{a+b}+\dfrac{b^{2}}{b+c}+\dfrac{a^{2}}{c+a}+\dfrac{c^{2}}{a+c}+\dfrac{c^{2}}{c+b}, a + b a 2 + a + b b 2 + b + c b 2 + c + a a 2 + a + c c 2 + c + b c 2 ,
which is ≥ ( 2 a + 2 b + 2 c ) 2 4 ( a + b + c ) , ≥\frac{(2a+2b+2c)^{2}}{4(a+b+c)}, ≥ 4 ( a + b + c ) ( 2 a + 2 b + 2 c ) 2 , so we are done. □ _\square □
Now you can try to prove the Nesbitt's inequality .
Nesbitt's Inequality
Let a , b , c ∈ R + a, b, c \in \mathbb{R}^+ a , b , c ∈ R + . Show that
a b + c + b a + c + c b + a ≥ 3 2 . \frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{b + a} \ge \frac{3}{2}. b + c a + a + c b + b + a c ≥ 2 3 .
SHOW SOLUTION
The numerator currently isn't a perfect square. Let's make it a perfect square:
a b + c + b a + c + c b + a = a 2 a b + a c + b 2 a b + b c + c 2 b c + a c . \frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{b + a} = \frac{ a^2 } { ab+ac} + \frac{ b^2 } { ab + bc } + \frac{ c^2 } { bc + ac } . b + c a + a + c b + b + a c = a b + a c a 2 + a b + b c b 2 + b c + a c c 2 .
We can then apply Titu's lemma to obtain
a 2 a b + a c + b 2 a b + b c + c 2 b c + a c ≥ ( a + b + c ) 2 2 ( a b + b c + c a ) . \frac{ a^2 } { ab+ac} + \frac{ b^2 } { ab + bc } + \frac{ c^2 } { bc + ac } \geq \frac{ (a+b+c) ^2 } { 2 ( ab+bc+ca) } . a b + a c a 2 + a b + b c b 2 + b c + a c c 2 ≥ 2 ( a b + b c + c a ) ( a + b + c ) 2 .
Now,
( a + b + c ) 2 2 ( a b + b c + c a ) ≥ 3 2 ⇔ ( a + b + c ) 2 ≥ 3 ( a b + b c + c a ) ⇔ a 2 + b 2 + c 2 ≥ a b + b c + c a ⇔ ( a − b ) 2 + ( b − c ) 2 + ( c − a ) 2 ≥ 0. \begin{array} { c r c l }
& \frac{ (a+b+c) ^2 } { 2 ( ab+bc+ca) } & \geq & \frac{3}{2} \\
\Leftrightarrow & (a+b+c)^2 & \geq & 3 (ab+bc+ca) \\
\Leftrightarrow & a^2 + b^2 + c^2 & \geq & ab + bc + ca \\
\Leftrightarrow & (a-b)^2+(b-c)^2+(c-a)^2 & \geq & 0.
\end{array} ⇔ ⇔ ⇔ 2 ( a b + b c + c a ) ( a + b + c ) 2 ( a + b + c ) 2 a 2 + b 2 + c 2 ( a − b ) 2 + ( b − c ) 2 + ( c − a ) 2 ≥ ≥ ≥ ≥ 2 3 3 ( a b + b c + c a ) a b + b c + c a 0 .
Note that if we tried to apply Titu's lemma directly, we end up with
a b + c + b a + c + c b + a ≥ ( a + b + c ) 2 2 ( a + b + c ) . \frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{b + a} \geq \frac{ \big( \sqrt{a} + \sqrt{b} + \sqrt{c} \big)^2 } { 2 ( a + b + c ) }. b + c a + a + c b + b + a c ≥ 2 ( a + b + c ) ( a + b + c ) 2 .
Then, the right-hand side actually has a maximum value of 3 2 \frac{3}{2} 2 3 , so we cannot proceed any further. □ _\square □
Another approach is to apply Titu's lemma to show that:
1 b + c + 1 c + a + 1 a + b ≥ ( 1 + 1 + 1 ) 2 2 ( a + b + c ) . \frac{1}{ b+c} + \frac{1}{ c+a} + \frac{1}{a+b} \geq \frac{ (1 + 1 + 1)^2 } { 2(a+b+c) }. b + c 1 + c + a 1 + a + b 1 ≥ 2 ( a + b + c ) ( 1 + 1 + 1 ) 2 .
Hence,
a + b + c b + c + a + b + c c + a + a + b + c a + b ≥ 9 2 . \frac{a+b+c}{ b+c} + \frac{a+b+c}{ c+a} + \frac{a+b+c}{a+b} \geq \frac{9}{2}. b + c a + b + c + c + a a + b + c + a + b a + b + c ≥ 2 9 .
Hence,
a b + c + b c + a + c a + b ≥ 9 2 − 3 = 3 2 . \frac{a}{ b+c} + \frac{b}{ c+a} + \frac{c}{a+b} \geq \frac{9}{2} - 3 = \frac{3}{2} . b + c a + c + a b + a + b c ≥ 2 9 − 3 = 2 3 .