Inequalities (Olympiad)hardFree

Maximum P in a Cyclic Inequality with a+b+c=1 | IOQM

JEE Maths question with a full step-by-step solution.

Question

a,b,c

are positive real numbers with

a+b+c=1

such that

\frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c}\ge\frac{P}{4},

then find the maximum possible value of

P

for which the inequality above is always true.

Solution

Answer: 69

Step 1: Split each numerator using

7+2x=5+2(1+x)

. For the first term,

7+2b=5+2(1+b)

, and similarly for the others, so

\frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c}=5\left[\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right]+2\left[\frac{1+b}{1+a}+\frac{1+c}{1+b}+\frac{1+a}{1+c}\right].

Step 2: Bound the first bracket using Titus's lemma (Bergström's inequality): for positive

x,y,z

\frac{p^2}{x}+\frac{q^2}{y}+\frac{r^2}{z}\ge\frac{(p+q+r)^2}{x+y+z}.

Taking

p=q=r=1

and using

a+b+c=1

\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge\frac{(1+1+1)^2}{(1+a)+(1+b)+(1+c)}=\frac{9}{3+(a+b+c)}=\frac{9}{4}.

Step 3: Bound the second bracket using the AM-GM inequality: for positive reals the sum is at least three times the cube root of their product. Since the cyclic product telescopes to

1

\frac{1+b}{1+a}+\frac{1+c}{1+b}+\frac{1+a}{1+c}\ge3\sqrt[3]{\frac{1+b}{1+a}\cdot\frac{1+c}{1+b}\cdot\frac{1+a}{1+c}}=3\sqrt[3]{1}=3.

Step 4: Combine the two bounds:

\frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c}\ge5\cdot\frac{9}{4}+2\cdot3=\frac{45}{4}+6=\frac{69}{4}.

Step 5: The given inequality

\text{(sum)}\ge\dfrac{P}{4}

holds for all admissible

a,b,c

exactly when

\dfrac{P}{4}\le\dfrac{69}{4}

, i.e.

P\le69

. The bound

\dfrac{69}{4}

is attained at

a=b=c=\dfrac13

(where each term equals

\dfrac{23}{4}

), so it cannot be improved. Hence the maximum value is

P=69.

Answer: 69

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