A contest-math inequality: $(a+b)^2+(a+b+4c)^2 \geq \frac{100abc}{a+b+c}$

Question

I completed my steps like this:

Let's prove our last inequality:

$$a^3 + 3 a^2 b + 5 a^2 c + 3 a b^2 + 12 a c^2 + b^3 + 5 b^2 c + 12 b c^2 + 8 c^3 \geq 40 a b c $$

Let, $$m=\frac ba, \qquad n=\frac ca$$

We have $$m^3+3m^2+3m+1+12n^2+5n+5m^2n+12mn^2+8n^3 \geq40 mn$$

Then,

$$m^3+3m^2+3m+1+12n^2+5n+5m^2n+12mn^2+8n^3=(m^3+8n^3+1)+(5n+5mn^2)+(3m^2+12n^2)+(3m+12mn^2) \geq 40mn$$

Finally we get,

$$\color {red} {\begin{cases} m^3+8n^3+1 \thinspace \geq \thinspace 6mn \\ 5n+5mn^2\thinspace \geq \thinspace 10mn \\ 3m^2+12n^2 \thinspace \geq \thinspace 12mn \\ 3m+12mn^2 \thinspace \geq \thinspace 12mn \end{cases}}\color {red}{\Longrightarrow} \\ \\ \\ \color{blue} \Longrightarrow \color{blue} {\boxed{m^3+3m^2+3m+1+12n^2+5n+5m^2n+12mn^2+8n^3 \geq 40 mn}} $$

Equality only holds for $m=1$ and $n=\dfrac 12$.

Q.E.D.

If $a,b,c$ are positive real numbers prove that the following inequality and find the equality condition.

$$(a+b)^2+(a+b+4c)^2 \geq \frac{100abc}{a+b+c}$$

This inequality was presented to the students on a official paper. I will write my attempts very short.

$$(a+b+c)((a+b)^2+(a+b+4c)^2)-100abc \geq 0 \\ 2 (a^3 + 3 a^2 b + 5 a^2 c + 3 a b^2 - 40 a b c + 12 a c^2 + b^3 + 5 b^2 c + 12 b c^2 + 8 c^3) \geq 0 \\ a^3 + 3 a^2 b + 5 a^2 c + 3 a b^2 + 12 a c^2 + b^3 + 5 b^2 c + 12 b c^2 + 8 c^3 \geq 40 a b c \\$$

We have,

$$a^3 + 3 a^2 b + 5 a^2 c + 3 a b^2 + 12 a c^2 + b^3 + 5 b^2 c + 12 b c^2 + 8 c^3 \geq \\ \geq 9 \sqrt[9]{a^3 \times 3 a^2 b \times 5 a^2 c \times 3 a b^2 \times 12 a c^2 \times b^3 \times 5 b^2 c \times 12 b c^2 \times 8 c^3}= \\ =9 \times (2^7 \times 3^4 \times 5^2 )^{\frac 19}\times abc\approx 35.955 abc \leq 40 abc $$

In other words,

$$\text{min} \left[ \frac {a^3 + 3 a^2 b + 5 a^2 c + 3 a b^2 + 12 a c^2 + b^3 + 5 b^2 c + 12 b c^2 + 8 c^3}{ a b c} \right]=9\times \sqrt[9]{259200} \approx 35.955 \leq40 $$

Answer 1

You've proven that $a^3 + 3 a^2 b + 5 a^2 c + 3 a b^2 + 12 a c^2 + b^3 + 5 b^2 c + 12 b c^2 + 8 c^3 \geq 35.955 abc$.

But you have not proven that $35.955abc \le a^3 + 3 a^2 b + 5 a^2 c + 3 a b^2 + 12 a c^2 + b^3 + 5 b^2 c + 12 b c^2 + 8 c^3 < 40 abc$ need ever occur.

For $a^3 + 3 a^2 b + 5 a^2 c + 3 a b^2 + 12 a c^2 + b^3 + 5 b^2 c + 12 b c^2 + 8 c^3 = 35.955 abc$ can only occur if:

$a^3 = a^2 b =5 a^2 c = 3 a b^2 = 12 a c^2= b^3 = 5 b^2 c = 12 b c^2= 8 c^3$ which is easy to show never occurs. [ If this occurs then $a=b$ and $12c=5b$ and $8c^3 = b^3$ so $8=\frac {12}{5}^3=(2\frac 25)^3$. ... Never occurs. ]

Remember the AM-GM theorem which says $A+B+C+ ... + I \ge 9\sqrt[9]{ABC....I}$ has equality holding only if $A=B =C=D.... =I$. If that equality can not hold, that $A+B+C_ ... + I$ must be strictly greater than $9\sqrt[9]{ABC....I}$ and proving $A+B+C+.... + I\ge 9\sqrt[9]{ABC....I}$ does not contradict $A+B+C+.... + I\ge K\sqrt[9]{ABC....I}$ for $K > 9$.

Answer 2

To demonstrate the inequality is wrong, all you need is a counterexample ... so given your analysis, can you come up with one? That should show it pretty quickly.

Answer 3

I think your inequality is right.

Let $c=x\sqrt{ab}.$

Thus, by AM-GM $$((a+b)^2+(a+b+4c)^2)(a+b+c)\geq(4ab+(2\sqrt{ab}+4c)^2)(2\sqrt{ab}+c).$$ Id est, it's enough to prove that $$(4+(2+4x)^2)(2+x)\geq100x$$ or $$(2x-1)^2(x+4)\geq0$$ and we are done.

Answer 4

theres another way i found which matches equality.

$(a+b)^2 + (a+b+4c)^2 \geq \frac{100abc}{a+b+c}$

By multiplying both side by a+b+c

$$2(a^2+b^2+8c^2+2ab+4bc+4ca)(a+b+c) \geq 100abc$$ Proving this is equivalence to proving $(a+b)^2 + (a+b+4c)^2 \geq \frac{100abc}{a+b+c}$

By AM-GM

$$2(a^2+b^2+4c^2+4c^2+ab+ab+2bc+2bc+2ca+2ca) \geq 20\sqrt[10]{2^8a^6b^6c^8} = 20\sqrt[5]{2^4a^3b^3c^4} $$ on the other side, $$\frac{a}{2}+\frac{a}{2}+\frac{b}{2}+\frac{b}{2}+c \geq 5\sqrt[5]{\frac{a^2b^2c}{2^4}}$$ By multiplying these 2 inequalities, $$2(a^2+b^2+4c^2+4c^2+ab+ab+2bc+2bc+2ca+2ca)(\frac{a}{2}+\frac{a}{2}+\frac{b}{2}+\frac{b}{2}+c )\geq 100\sqrt[5]{a^5b^5c^5} = 100abc$$ The equality is when a = b = 2c

Q.E.D.