Random reals.

Algebra Level 4

a , b , c a,b,c are positive real numbers such that 1 b + c + 1 c + a + 1 a + b = 1 \dfrac{1}{b+c} + \dfrac{1}{c+a} + \dfrac{1}{a+b} = 1 . If the minimum value of 1 a 5 + 1 b 5 + 1 c 5 \dfrac{1}{a^5} + \dfrac{1}{b^5} + \dfrac{1}{c^5} can be written as p q \dfrac{p}{q} where p p and q q are positive integers with gcd(p,q) = 1 \text{gcd(p,q)}=1 , find ( p + q ) (p+q)


The answer is 113.

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Sagnik Saha by Sagnik Saha , 23, India

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3 solutions

Kartik Sharma
Dec 29, 2014

Nice problem! Well, I did it in a rather 'wrong ' way(I don't know if this is correct or not).

1 b + c + 1 c + a + 1 a + b = 1 \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} = 1

Multiplying a + b + c a+b+c ,

a b + c + b c + a + c b + a + 3 = a + b + c \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{b+a} + 3 = a+b+c

Equating the minimum of both the sides and then using Nesbitt's inequality,

3 2 + 3 = 3 a b c 3 \frac{3}{2} + 3 = 3\sqrt[3]{abc}

a b c = 3 2 3 abc = {\frac{3}{2}}^{3}

Now, using AM-GM in 2nd equation,

1 a 5 + 1 a 5 + 1 a 5 ≥ 3 1 a b c 5 3 \frac{1}{{a}^{5}} + \frac{1}{{a}^{5}} + \frac{1}{{a}^{5}} \geq 3\sqrt[3]{\frac{1}{{abc}^{5}}}

Putting value of abc here, we get

≥ 2 5 3 4 \geq \frac{{2}^{5}}{{3}^{4}}

≥ 32 81 \geq \frac{32}{81}

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Aaaaa Bbbbb
Apr 16, 2014

It is easy to prove that: 1 ( a − t ) 5 + 1 ( a + t ) 5 > 2 a 5 : t > 0 , a > 0 \frac{1}{(a-t)^{5}} + \frac{1}{(a+t)^{5}} > \frac{2}{a^{5}}: t>0, a>0 So 1 a 5 + 1 b 5 + 1 c 5 \frac{1}{a^{5}}+\frac{1}{b^{5}}+\frac{1}{c^{5}} is maximum when: a = b = c = 3 2 a=b=c=\frac{3}{2} M I N ( 1 a 5 + 1 b 5 + 1 c 5 ) = 32 81 = p q ⇒ p + q = 113 MIN(\frac{1}{a^{5}}+\frac{1}{b^{5}}+\frac{1}{c^{5}})=\frac{32}{81}=\frac{p}{q} \Rightarrow p+q=\boxed{113}

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Introduce x = 1 a x=\dfrac{1}{a} , y = 1 b y=\dfrac{1}{b} , z = 1 c z=\dfrac{1}{c} .

Then 1 a + b + 1 b + c + 1 c + a \dfrac{1}{a+b} + \dfrac{1}{b+c} + \dfrac{1}{c+a} becomes x y x + y + y z y + z + z x z + x \dfrac{xy}{x+y} + \dfrac{yz}{y+z} + \dfrac{zx}{z+x} . We know ( x + y ) 2 ≥ 4 x y ⟹ ( x + y ) 4 ≥ x y x + y (x+y)^2 \geq 4xy \implies \dfrac{(x+y)}{4} \geq \dfrac{xy}{x+y} . Continuing this way and adding we get 2 ⋅ x + y + z 4 ≥ x y x + y + y z y + z + z x z + x = 1 2 \cdot \dfrac{x+y+z}{4} \geq \dfrac{xy}{x+y} + \dfrac{yz}{y+z} + \dfrac{zx}{z+x} = 1 . Therefore we have ( x + y + z ) ≥ 2 (x+y+z) \geq 2 with equality when x = y = z = 2 3 x=y=z=\frac{2}{3} . Now, We use power mean inequality to arrive at, x 5 + y 5 + z 5 3 ≥ ( x + y + z ) 5 3 5 ⟹ x 5 + y 5 + z 5 ≥ ( x + y + z ) 5 3 4 ≥ 2 5 3 5 = 32 81 \dfrac{x^5 + y^5 + z^5}{3} \geq \dfrac{(x+y+z)^5}{3^5} \implies x^5 + y^5 + z^5 \geq \dfrac{(x+y+z)^5}{3^4} \geq \dfrac{2^5}{3^5} = \boxed{\dfrac{32}{81}} . Both the inequalitites satisfy the inequality case when x = y = z = 2 / 3 x=y=z=2/3 and it indeed gives the value equal to 32 81 \dfrac{32}{81} .

Sagnik Saha - 7 years, 1 month ago

The minimum value of the second equation would only be reached if a = b = c a=b=c . Thus, the first equation would become 3 2 a \frac{3}{2a} = 1

Thus, a=b=c= 3 2 \frac{3}{2}

The second equation boils down to 2 5 3 4 \frac{2^{5}}{3^{4}} = 32 81 \frac{32}{81}

Thus the answers is 32+81= 113 \boxed{113}

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