inequality questionfest

Algebra Level 2

Given positive reals $a, b, c$ such that $a + b + c = 4$ , what is the maximum value of

$a^2bc + ab^2c + abc^2?$

\frac{100}{63}

4

\frac{55}{23}

\frac{256}{27}

2 solutions

Madhav Srirangan
Dec 30, 2014

Apply AM-GM inequality on $a,b,c$ $\frac{a+b+c}{3}\geq \sqrt[3]{abc}\\4 /3\geq \sqrt[3]{abc}\\ \frac{64}{27}\geq abc$ Therefore for maximum value take $abc=\dfrac{64}{27}$ $a^2bc+b^2ca+c^2ab=abc(a+b+c)\\=\frac{64}{27}\times 4\\ =\boxed{\frac{256}{27}}$

but this is giving maximum value. .you should have asked maximum value

Subhrajyoti Sinha - 6 years, 5 months ago

Vishal S
Dec 31, 2014

By A.M.-G.M. inequality a+b+c / 3 ≥ (abc)^1/3

But given a+b+c=4

therefore a+b+c / 3 ≥ (abc)^1/3 =>4/3 ≥ (abc)^1/3 =>64/27 ≥ abc ---->(1)

Now a^2bc+b^2ac+c^2ab can be written as abc(a+b+c) =>4abc ---->(2)

(1)=>The largest value of abc is 16/27. By substituting abc in (2)

We get abc(a+b+c)</= 4 x 64/27=256/7

therefore the maximum value of a^2bc+b^2ac+c^2ab is 256/7

The above solution gives the maximum value.Please edit the question

it is >= means min 256/7 or more than that therefore min value only baby.......

madhav srirangan - 6 years, 5 months ago

Can you fix the inequality in the third line? It should be $64 / 27 \geq abc$ . please fix the follow through errors too.

Can you fix the inequality sign and the denominator in the 3rd last line? It should be

$abc ( a + b + c ) \leq 4 \times 64/27 = 256 / 27$

Calvin Lin Staff - 6 years, 5 months ago

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