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Question: For positive real numbers \[a,b\] and \[c\] such that \[a + b + c = p\], which one holds true? ...

For positive real numbers a,ba,b and cc such that a+b+c=pa + b + c = p, which one holds true?

A. (pa)(pb)(pc)827p3\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right) \leqslant \dfrac{8}{{27}}{p^3}
B. (pa)(pb)(pc)8abc\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right) \geqslant 8abc
C. bca+cab+abcp\dfrac{{bc}}{a} + \dfrac{{ca}}{b} + \dfrac{{ab}}{c} \leqslant p
D. None of the above

Explanation

Solution

- Hint: The arithmetic mean of the given positive numbers is greater or equal to the geometric mean of the positive numbers. The arithmetic mean and geometric mean is equal if each and every element is identical to each other.

Complete step-by-step solution -
(i). The given numbers a,bandca,b\,\,{\text{and}}\,\,c are positive real numbers, therefore,
Arithmetic MeanGeometric Mean{\text{Arithmetic Mean}} \geqslant {\text{Geometric Mean}}
Now, for aa andbb,
a+b2ab......(1)\dfrac{{a + b}}{2} \geqslant \sqrt {ab} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 1 \right)
For bb andcc,
b+c2bc......(2)\dfrac{{b + c}}{2} \geqslant \sqrt {bc} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 2 \right)
Similarly, for aa andcc,
a+c2ac......(3)\dfrac{{a + c}}{2} \geqslant \sqrt {ac} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 3 \right)
Now, multiply equation (1), (2) and (3) as shown below.

(a+b)(b+c)(c+a)8a2b2c2 (a+b)(b+c)(c+a)8abc (pa)(pb)(pc)8abc(a+b+c=p)  \,\,\,\,\,\,\,\dfrac{{\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)}}{8} \geqslant \sqrt {{a^2}{b^2}{c^2}} \\\ \Rightarrow \left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) \geqslant 8abc \\\ \Rightarrow \left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right) \geqslant 8abc\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {a + b + c = p} \right) \\\

Therefore, option (B) is correct.
(ii). Similarly apply Arithmetic MeanGeometric Mean{\text{Arithmetic Mean}} \geqslant {\text{Geometric Mean}} over(pa),(pb)and(pc)\left( {p - a} \right),\left( {p - b} \right)\,\,{\text{and}}\,\,\left( {p - c} \right).

3p(a+b+c)3[(pa)(pb)(pc)]13 2p3[(pa)(pb)(pc)]13 (pa)(pb)(pc)8p327  \,\,\,\,\,\,\,\,\dfrac{{3p - \left( {a + b + c} \right)}}{3} \geqslant {\left[ {\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right)} \right]^{\dfrac{1}{3}}} \\\ \Rightarrow \dfrac{{2p}}{3} \geqslant {\left[ {\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right)} \right]^{\dfrac{1}{3}}} \\\ \Rightarrow \left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right) \leqslant \dfrac{{8{p^3}}}{{27}} \\\

Therefore, the option (A) is correct.
(iii). Now, consider bca,cabandabc\dfrac{{bc}}{a},\dfrac{{ca}}{b}\,\,{\text{and}}\,\,\dfrac{{ab}}{c} be positive real numbers, therefore,
Arithmetic MeanGeometric Mean{\text{Arithmetic Mean}} \geqslant {\text{Geometric Mean}}
For, bcaandcab\dfrac{{bc}}{a}\,\,{\text{and}}\,\,\dfrac{{ca}}{b},

12(bca+cab)c bca+cab2c......(4)  \,\,\,\,\,\,\,\dfrac{1}{2}\left( {\dfrac{{bc}}{a} + \dfrac{{ca}}{b}} \right) \geqslant c \\\ \Rightarrow \dfrac{{bc}}{a} + \dfrac{{ca}}{b} \geqslant 2c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 4 \right) \\\

For, cabandabc\dfrac{{ca}}{b}\,\,{\text{and}}\,\,\dfrac{{ab}}{c},
cab+abc2a......(5)\dfrac{{ca}}{b} + \dfrac{{ab}}{c} \geqslant 2a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 5 \right)
For, bcaandabc\dfrac{{bc}}{a}\,\,{\text{and}}\,\,\dfrac{{ab}}{c},
bca+abc2b......(6)\dfrac{{bc}}{a} + \dfrac{{ab}}{c} \geqslant 2b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 6 \right)
Now, add equation (4),(5) and (6).

2(bca+cab+abc)2(a+b+c) bca+cab+abca+b+c bca+cab+abcp(a+b+c=p)  \,\,\,\,\,\,\,2\left( {\dfrac{{bc}}{a} + \dfrac{{ca}}{b} + \dfrac{{ab}}{c}} \right) \geqslant 2\left( {a + b + c} \right) \\\ \Rightarrow \dfrac{{bc}}{a} + \dfrac{{ca}}{b} + \dfrac{{ab}}{c} \geqslant a + b + c \\\ \Rightarrow \dfrac{{bc}}{a} + \dfrac{{ca}}{b} + \dfrac{{ab}}{c} \geqslant p\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {a + b + c = p} \right) \\\

Therefore, the option (C) is not correct.

Thus, option (A) and (B) are correct.

Note: The arithmetic mean of two numbers is the ratio of the sum of the numbers to the total numbers. The geometric mean of two numbers is the square root of the products of the two numbers.