Prove that: $2(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge\sqrt{5ab+4ac}+\sqrt{5bc+4ba}+\sqrt{5ca+4cb}$

Question

Problem: For $a,b,c\ge0: ab+bc+ca>0.$ Prove that: $$2(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge\sqrt{5ab+4ac}+\sqrt{5bc+4ba}+\sqrt{5ca+4cb}$$

Recently, i have seen a post on AoPS link

My approach: After squaring both side, i get: $$2(a^2+b^2+c^2)+\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})+2(a+b+c)(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\ge\sum{\sqrt{(5ab+4ac)(5bc+4ba)}}$$ Since: $a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}$. It is desired to prove: $$2(a^2+b^2+c^2)+5\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})+2(ab+bc+ca)\ge\sum{\sqrt{(5ab+4ac)(5bc+4ba)}}$$

But it is more complicated. I hope we can find a good approach for problem. Thanks!

Answer 1

Remarks: We use computer to motivate the solution. I think there is better way to prove (2).

We may use isolated fudging.

It suffices to prove that $$ [2(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}]\cdot \frac{4\sqrt{ab} + 5\sqrt{bc}}{9(\sqrt{ab} + \sqrt{bc} + \sqrt{ca})} \ge \sqrt{4ab + 5bc}. \tag{1} $$ (Note: Take cyclic sum on (1), we get the desired inequality.)

Let $x = \sqrt{ab}, y = \sqrt{bc}, z = \sqrt{ca}$. (1) is written as $$\left(\frac{2zx}{y} + \frac{2xy}{z} + \frac{2yz}{x} + x + y + z\right)\cdot \frac{4x + 5y}{9(x + y + z)} \ge \sqrt{4x^2 + 5y^2}. \tag{2}$$

We have \begin{align*} &\left(\frac{2zx}{y} + \frac{2xy}{z} + \frac{2yz}{x} + x + y + z\right)\cdot \frac{4x + 5y}{9(x + y + z)}\\ ={}& \frac{(5x^2 + 2xy + 5y^2)(4x + 5y)}{9(x + y)^2} + \frac{2(4x + 5y)(x^2y - x^2z + xy^2 - y^2z)^2}{9xyz(x + y + z)(x + y)^2}\\ \ge{}& \frac{(5x^2 + 2xy + 5y^2)(4x + 5y)}{9(x + y)^2}. \end{align*}

It suffices to prove that $$\frac{(5x^2 + 2xy + 5y^2)(4x + 5y)}{9(x + y)^2} \ge \sqrt{4x^2 + 5y^2}$$ which is true using $$\mathrm{LHS}^2 - \mathrm{RHS}^2 = \frac{4(19x^4 + 44x^3y + 54x^2y^2 + 80xy^3 + 55y^4)(x - y)^2}{81(x + y)^4} \ge 0.$$

We are done.

Answer 2

Proof.

By denoting $a=x^2;b=y^2;c=z^2$, we will prove $$2\sum_{cyc} x^2 + \sum_{cyc} yz \ge \sum_{cyc} x\sqrt{5z^2+4y^2}.$$ By using Cauchy-Schwarz $$\sum_{cyc} x\sqrt{5z^2+4y^2}=\sum_{cyc} \sqrt{x}\sqrt{5xz^2+4xy^2}\le \sqrt{\sum_{cyc} x .\sum_{cyc} (5xz^2+4xy^2)}.$$ Hence, it suffices to prove $$\left(2\sum_{cyc} x^2 + \sum_{cyc} yz \right)^2\ge \sum_{cyc} x .\sum_{cyc} (5xz^2+4xy^2),$$ which is equivalent to $$4\sum x^4 - \sum x^3y - 3\sum x^2yz \ge 0 .$$The last inequality is true by AM-GM.

Indeed, $$x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\ge xyz(x+y+z),$$ and $$3.x^4+y^4\ge 4x^3y;3.y^4+z^4\ge 4y^3z;3.z^4+x^4\ge 4z^3x.$$ Hence, the result follows. Equality holds at $a=b=c.$

Answer 3

We need to prove that: $$\sum_{cyc}(2a^2+ab)\geq\sum_{cyc}a\sqrt{5b^2+4c^2},$$ where $a$, $b$ and $c$ are non-negatives.

Indeed, after squaring of the both sides we need to prove that: $$\sum_{cyc}(4a^4+8a^2b^2+a^2b^2+2a^2bc+4a^3b+4a^3c+4a^2bc-9a^2b^2)\geq2\sum_{cyc}ab\sqrt{5b^2+4c^2)(5c^2+4a^2)}$$ or $$\sum_{cyc}(2a^4+2a^3b+2a^3c+3a^2bc)\geq\sum_{cyc}ab\sqrt{5b^2+4c^2)(5c^2+4a^2)}$$ or $$(a+b+c)(2(a^3+b^3+c^3)+3abc)\geq\sum_{cyc}ab\sqrt{5b^2+4c^2)(5c^2+4a^2)}.$$ Now, by C-S and Rearrangement $$\sum_{cyc}ab\sqrt{5b^2+4c^2)(5c^2+4a^2)}\leq\sqrt{\sum_{cyc}ab\sum_{cyc}ab(5b^2+4c^2)(5c^2+4a^2)}=$$ $$=\sqrt{\sum_{cyc}ab\sum_{cyc}(20a^3b^3+20a^4bc+25a^3b^2c+16a^3c^2b)}\leq$$ $$\leq\sqrt{\sum_{cyc}ab\sum_{cyc}(20a^3b^3+29a^4bc+16a^3b^2c+16a^3c^2b)}.$$ Id est, it's enough to prove that: $$(a+b+c)^2(2(a^3+b^3+c^3)+3abc)^2\geq(ab+ac+bc)\sum_{cyc}(20a^3b^3+29a^4bc+16a^3b^2c+16a^3c^2b).$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$, $abc=w^3$ and $u^2=tv^2.$

Thus, $t\geq1$ and we need to prove that: $$9u^2(54u^3-54yv^2+6w^3+3w^3)^2\geq$$ $$\geq3v^2(20(27v^6-27uv^2w^3+3w^6)+29(27u^3w^3-27uv^2w^3+3w^6)+16(9uv^2-3w^3)w^3)$$ or $$27u^2(6u^3-6uv^2+w^3)^2\geq v^2(60v^6+87u^3w^3-131uv^2w^3+11w^6)$$ or $$972u^8-1944u^6v^2+972u^4v^4-60v^8+(324u^5-411u^3v^2+131uv^4)w^3+(27u^2-11v^2)w^6\geq0,$$ which is true by Schur ($w^3\geq4uv^2-3u^3$) twice: $$972u^8-1944u^6v^2+972u^4v^4-60v^8+(324u^5-411u^3v^2+131uv^4)w^3+(27u^2-11v^2)w^6\geq$$ $$\geq972u^8-1944u^6v^2+972u^4v^4-60v^8+(324u^5-411u^3v^2+131uv^4)w^3+(27u^2-11v^2)(4uv^2-3u^3)w^3=$$ $$=972u^8-1944u^6v^2+972u^4v^4-60v^8+(243u^5-270u^3v^2+87uv^4)w^3\geq$$ $$\geq972u^8-1944u^6v^2+972u^4v^4-60v^8+(243u^5-270u^3v^2+87uv^4)(4uv^2-3u^3)=$$ $$=3v^8t(t-1)(81t^3+27t^2-96t+20)\geq0.$$

Answer 4

Alternative proof.

By multiplying $a+b+c$ for both side, we will prove $$2(a+b+c)^2+(a+b+c)\sum_{cyc}\sqrt{bc}\ge (a+b+c)\sum_{cyc}\sqrt{5ab+4ac}.$$

From now, we can apply Michael Rozenberg's CBS using.

Indeed, $$(a+b+c)\sqrt{5ab+4ac}=\sqrt{[(b+c-a)^2+4a(b+c)].[ab+4a(b+c)]}\ge \sqrt{ab}(b+c-a)+4a(b+c).$$ It implies$$(a+b+c)\left(\sqrt{5ab+4ac}+\sqrt{ab}\right)\ge 2(b+c)(2a+\sqrt{ab}),$$ $$\iff \frac{a+b+c}{2a+\sqrt{ab}}\ge \frac{2(b+c)}{\sqrt{5ab+4ac}+\sqrt{ab}},$$ $$\iff \frac{a(a+b+c)^2}{a+\dfrac{\sqrt{ab}}{2}}+\sqrt{ab}(a+b+c)\ge (a+b+c)\sqrt{5ab+4ac}.\tag{*}$$

Notice that: $\dfrac{2(b+c)}{\sqrt{5ab+4ac}+\sqrt{ab}}=\dfrac{\sqrt{5ab+4ac}-\sqrt{ab}}{2a}.$

Taking cyclic sum on $(*),$ we obtain$$ (a+b+c)\sum_{cyc}\sqrt{5ab+4ac}\le (a+b+c)^2.\sum_{cyc}\frac{a}{a+\dfrac{\sqrt{ab}}{2}}+(a+b+c)\sum_{cyc}\sqrt{bc}.$$ Id est, it is enough to prove that$$\sum_{cyc}\frac{a}{a+\dfrac{\sqrt{ab}}{2}}\le 2,$$ or $$\frac{\sqrt{a}}{2\sqrt{a}+\sqrt{b}}+\frac{\sqrt{b}}{2\sqrt{b}+\sqrt{c}}+\frac{\sqrt{c}}{2\sqrt{c}+\sqrt{a}}\le 1,$$ which is easy by Cauchy-Schwarz.

We end proof here. Equality holds at $a=b=c.$