Prove $3a+4(b+c)+\sqrt[3]{abc}\ge 2\left( \sqrt{b\left(4b+5a\right) }+ \sqrt{c\left(4c+5a\right) }\right).$

Question

For any $a,b,c\ge 0$ prove that $$3a+4(b+c)+\sqrt[3]{abc}\ge 2\left( \sqrt{b\left(4b+5a\right) }+ \sqrt{c\left(4c+5a\right) }\right).$$

Equality holds at $a=b=c=1$ or $a=0$

I tried to use AM as $$6\sqrt{b\left(4b+5a\right) }=2\sqrt{9b\left(4b+5a\right) }\le 13b+5a$$ $$6\sqrt{c\left(4c+5a\right) }=2\sqrt{9c\left(4c+5a\right) }\le 13c+5a$$ and we need to prove $$9a+12(b+c)+3\sqrt[3]{abc}\ge 13b+5a+13c+5a$$ or $$3\sqrt[3]{abc}\ge a+b+c$$which is reverse since by AM-GM $3\sqrt[3]{abc}\le a+b+c.$

Is there a better approach? Any ideas and comments is welcome.

Answer 1

This is not an answer, might be too long for a comment. It is definitely not elegant but probably doable if one tries hard. : )

Let us assume $a > 0$ otherwise the equality holds, and let $u=\frac{b}{a}$, $v=\frac{c}{a}$, then the problem is equivalent to

$$f(u,v) := 3 + 4(u+ v) + u^{1/3} v^{1/3} - 2 u^{1/2} (4u+5)^{1/2} - 2 v^{1/2} (4v+5)^{1/2}\ge 0$$

and one can exclude the possibility of a local minimizer on the boundary, then solve

$$f_u = \frac{1}{3} u^{-2/3} v^{1/3} - \frac{(2\sqrt{u} - \sqrt{4u+5})^2}{\sqrt{u(4u+5)}} = 0$$

$$f_v = \frac{1}{3} v^{-2/3} u^{1/3} - \frac{(2\sqrt{v} - \sqrt{4v+5})^2}{\sqrt{v(4v+5)}} = 0$$

which gives $v = F(u), u = F(v)$, where $$F(x) =27 \left( \frac{(2\sqrt{x} - \sqrt{4x+5})^2}{\sqrt{x(4x+5)}} \right)^3 x^{2} = 27\left( \frac{25}{\sqrt{x(4x+5)}(2\sqrt{x} + \sqrt{4x+5})^2} \right)^3 x^{2}=27 \left( \frac{25}{\sqrt{(4x+5)}(2\sqrt{x} + \sqrt{4x+5})^2} \right)^3 x^{1/2},$$ then all we need is the fixed point of $F\circ F$, it seems there is no quick way for that.

Answer 2

We need to prove that: $$(3a^3+4b^3+4c^3+abc)^2\geq4\left(\sqrt{b^3(4b^3+5a^3)}+\sqrt{c^3(4c^3+5a^3)}\right)^2$$ or $$9a^6+4(b^3+c^3)a^3+32b^3c^3+2abc(3a^3+4b^3+4c^3)+a^2b^2c^2\geq8\sqrt{b^3c^3(5a^3+4b^3)(5a^3+4c^3)}$$ or $$81a^{11}+108bca^9+72(b^3+c^2)a^8+54b^2c^2a^7+192bc(b^3+c^3)a^6+4(4b^6-245b^3c^3+4c^6)a^5+104b^2c^2(b^3+c^3)a^4+bc(64b^6+513b^3c^3+64c^3)a^3-1008b^3c^3(b^3+c^3)a^3+64b^2c^2(b^6+3b^3c^3+b^6)a+512b^4c^4(b^3+c^3)\geq0,$$ which is true by AM-GM.

Indeed, let $a=x\sqrt{bc}.$

Thus, $$81a^{11}+108bca^9+72(b^3+c^2)a^8+54b^2c^2a^7+192bc(b^3+c^3)a^6+4(4b^6-245b^3c^3+4c^6)a^5+104b^2c^2(b^3+c^3)a^4+bc(64b^6+513b^3c^3+64c^3)a^3-1008b^3c^3(b^3+c^3)a^2+64b^2c^2(b^6+3b^3c^3+b^6)a+512b^4c^4(b^3+c^3)=$$ $$=(b^3+c^3)(72a^8+96bca^6+8(b^3+c^3)a^5+104b^2c^2a^4+40bc(b^3+c^3)-1008b^3c^3a^2+64b^2c^2(b^3+c^3)a+512b^4c^4)+$$ $$+81a^{11}+108bca^9+54b^2c^2a^7+96bc(b^3+c^2)a^6+4(2b^6-249b^3c^3+2c^6)a^5+bc(24b^6+433b^3c^3+24c^6)a^3+64b^5c^5a\geq$$ $$\geq(b^3+c^3)(72a^8+96bca^6+16\sqrt{b^3c^3}a^5+104b^2c^2a^4+80bc\sqrt{b^3c^3}-1008b^3c^3a^2+128b^2c^2\sqrt{b^3c^3}a+512b^4c^4)+$$ $$+81a^{11}+108bca^9+54b^2c^2a^7+192bc\sqrt{b^3c^3}a^6-980b^3c^3a^5+481b^4c^4a^3+64b^5c^5a=$$ $$=8b^4c^4(b^3+c^3)(9x^8+12x^6+2x^5+13x^4+10x^3-126x^2+16x+64)+$$ $$+b^5c^5a(81x^{10}+108x^8+54x^6+192x^5-980x^4+481x^2+64)\geq$$ $$\geq8b^4c^4(b^3+c^3)\left(126\sqrt[126]{\left(x^8\right)^9\cdot\left(x^6\right)^{12}\cdot\left(x^5\right)^2\cdot\left(x^4\right)^{13}\cdot\left(x^3\right)^{10}\cdot x^{16}\cdot1^{64}}-126x^2\right)+$$ $$+b^5c^5a\left(980\sqrt[980]{\left(x^{10}\right)^{81}\cdot\left(x^8\right)^{108}\cdot\left(x^6\right)^{54}\cdot\left(x^5\right)^{192}\cdot\left(x^2\right)^{481}\cdot1^{64}}-980x^4\right)=0.$$

Answer 3

Sketch of a proof.

If $a = 0$, clearly the desired inequality is true.

If $a > 0$, WLOG, assume that $a = 1$.

The desired inequality is written as $$\left(3 + 4(b + c) + \sqrt[3]{bc}\right)^2 \ge 4b(4b + 5) + 4c(4c + 5) + 8\sqrt{bc}\sqrt{(4b + 5)(4c + 5)}. \tag{1}$$

Let $p = b + c, q = bc$. We have $p^2 \ge 4q$.

(1) is written as \begin{align*} f(p) &:= \left(3 + 4p + \sqrt[3]{q}\right)^2 - 20p - 16p^2 + 32q - 8\sqrt{q(25 + 20p + 16q)}\\ &\ge 0. \end{align*} We have \begin{align*} f'(p) &= 4 + 8\sqrt[3]{q} - \frac{80q}{\sqrt{q(25 + 20p + 16q)}}\\[6pt] &\ge 4 + 8\sqrt[3]{q} - \frac{80q}{\sqrt{q(25 + 20\cdot \sqrt{4q} + 16q)}}\\[6pt] &> 0 \end{align*} where we use $p \ge \sqrt{4q}$. (Note: For the last inequality, let $q = u^6$.)

Also, we have $f(\sqrt{4q}) \ge 0$ (let $q = u^6$). Thus, $f(p) \ge 0$ for all $p \ge \sqrt{4q}$.

We are done.

Answer 4

Proof.

We divide the inequality into two cases.

$\bullet: a=0$ it's obvious.

$\bullet: a>0$ Divide both side by $a$ $$3+4\left(\frac{b}{a}+\frac{c}{a}\right)+\sqrt[3]{\frac{bc}{a^2}}\ge 2\left(\sqrt{\frac{4b^2}{a^2}+\frac{5b}{a}}+\sqrt{\frac{4c^2}{a^2}+\frac{5c}{a}}\right).$$Now, let $x=\dfrac{b}{a};y=\dfrac{c}{a}$ and we'll prove $$4(x+y)+\sqrt[3]{xy}+3\ge 2\left(\sqrt{4x^2+5x}+\sqrt{4y^2+5y}\right); \forall x,y>0. \tag{*}$$ Rewrite $(*)$ as $$\sqrt[3]{xy}+3\ge 2\left(\sqrt{4x^2+5x}-2x+\sqrt{4y^2+5y}-2y\right)$$ or$$\sqrt[3]{xy}+3\ge 10\left(\frac{x}{\sqrt{4x^2+5x}+2x}+\frac{y}{\sqrt{4y^2+5y}+2y}\right)$$or$$\sqrt[3]{xy}+3\ge 10\left(\frac{1}{\sqrt{4+\dfrac{5}{x}}+2}+\frac{1}{\sqrt{4+\dfrac{5}{y}}+2}\right).$$ By letting$(x,y)\rightarrow \left(\dfrac{1}{x},\dfrac{1}{y}\right)$ it turns out$$\frac{1}{\sqrt[3]{xy}}+5\ge 10\left(\frac{1}{\sqrt{5x+4}+2}+\frac{1}{\sqrt{5y+4}+2}+\frac{1}{5}\right).$$ By using CBS inequality with $t=\sqrt[3]{xy}>0$ $$(5t+1)\left(\sqrt{5x+4}+2\right)=10t+2+\sqrt{\left((5t-1)^2+20t\right)(4+5x)}\ge 20t+10\sqrt{xt},$$ it implies$$\frac{1}{\sqrt{5x+4}+2}\le \frac{5t+1}{10\sqrt{t}\left(2\sqrt{t}+\sqrt{x}\right)}.$$ We obtain$$\frac{10}{\sqrt{4+5x}+2}\le \frac{5+\dfrac{1}{\sqrt[3]{xy}}}{2+\sqrt[6]{\dfrac{x^2}{y}}};\frac{10}{\sqrt{4+5x}+2}\le \frac{5+\dfrac{1}{\sqrt[3]{xy}}}{2+\sqrt[6]{\dfrac{y^2}{x}}}.$$ Also$$ 2\le \frac{5+\dfrac{1}{\sqrt[3]{xy}}}{2+\dfrac{1}{\sqrt[6]{xy}}} \iff \left(\dfrac{1}{\sqrt[6]{xy}}-1\right)^2\ge 0.$$ It remains to prove$$\frac{1}{2+\sqrt[6]{\dfrac{x^2}{y}}}+\frac{1}{2+\sqrt[6]{\dfrac{y^2}{x}}}+\frac{1}{2+\dfrac{1}{\sqrt[6]{xy}}}\le 1.$$ By denoting $m=\sqrt[6]{\dfrac{x^2}{y}}; n=\sqrt[6]{\dfrac{y^2}{x}}; p=\dfrac{1}{\sqrt[6]{xy}}$ then $m,n,p>0: mnp=1.$ We get a well-known inequality $$\frac{1}{m+2}+\frac{1}{n+2}+\frac{1}{p+2}\le 1.$$ Hence, the proof is done.

Answer 5

Alternative proof.

We'll use AM-GM which is inspired by Michael Rozenberg's idea.

Similarly to my previous answer, we'll prove$$4(x+y)+\sqrt[3]{xy}+3\ge 2\left(\sqrt{4x^2+5x}+\sqrt{4y^2+5y}\right); \forall x,y>0. $$ We may use AM-GM as \begin{align*} 2\sqrt{4x^2+5x}&\le \frac{4x^2+2x\sqrt[3]{xy}-2x\sqrt[3]{xy}+5x}{2x+\sqrt[3]{xy}}+2x+\sqrt[3]{xy}\\&=4x+\sqrt[3]{xy}+\frac{x}{2x+\sqrt[3]{xy}}\left(5-2\sqrt[3]{xy}\right). \end{align*} It's enough to prove$$3-\sqrt[3]{xy}+\left(2\sqrt[3]{xy}-5\right)\left(\frac{x}{2x+\sqrt[3]{xy}}+\frac{y}{2y+\sqrt[3]{xy}}\right)\ge 0 (*).$$ Rewrite $(*)$ as$$M+\left(5-2\sqrt[3]{xy}\right)\left(1-\frac{1}{2+\sqrt[3]{xy}}-\frac{1}{2+\sqrt[3]{\dfrac{y}{x^2}}}-\frac{1}{2+\sqrt[3]{\dfrac{y}{x^2}}}\right)\ge 0.$$where$$M=3-\sqrt[3]{xy}+2\sqrt[3]{xy}-5+\frac{5-2\sqrt[3]{xy}}{2+\sqrt[3]{xy}}=\frac{\left(1-\sqrt[3]{xy}\right)^2}{2+\sqrt[3]{xy}}\ge 0.$$ It is easy to prove $$1-\frac{1}{2+\sqrt[3]{xy}}-\frac{1}{2+\sqrt[3]{\dfrac{y}{x^2}}}-\frac{1}{2+\sqrt[3]{\dfrac{y}{x^2}}}\ge0 .$$ Thus, $(*)$ is true when$5-2\sqrt[3]{xy}\ge 0.$

In case $5-2\sqrt[3]{xy}< 0$ we use AM-GM$$6+8(x+y)+2\sqrt[3]{xy}> 1+\left(10+8(x+y)\right)\ge 4\left( \sqrt{4x^2+5x}+ \sqrt{4y^2+5y}\right).$$

The proof is completed.