Резюме. The paper considers interesting proofs of two algebraic inequalities.

Ключови думи: inequality; radical; generalisation

First, we will give the proof of the next algebraic inequality with radicals:

(1) \[ \sqrt{x^{2}+3}+\sqrt{y^{2}+3}+\sqrt{x y+3} \geq 6 ;(x, y \geq 0, x+y=2) \]

Proof: We will prove the two inequalities:

(2) \[ \sqrt{x^{2}+3}+\sqrt{y^{2}+3} \geq \sqrt{19-3 x y} \]

and

(3) \[ \sqrt{19-3 x y}+\sqrt{x y+3} \geq 6 \]

where \(x, y \geq 0\) and \(x+y=2\).

\(1^{0}\). We have \(0 \leq x y=t \leq 1\) because \(\tfrac{x+y}{2} \geq \sqrt{x y}\), i.e. \(1 \geq \sqrt{x y} \Rightarrow 0 \leq x y \leq 1\). On t the other hand

\[ x^{2}+y^{2}=(x+y)^{2}-2 x y=4-2 t \]

After squaring, we get from (2):

\[ \begin{aligned} & x^{2}+3+y^{2}+3+2 \sqrt{x^{2}+3} \cdot \sqrt{y^{2}+3} \geq 19-3 x y \\ \Leftrightarrow & (x+y)^{2}-2 x y+6+2 \sqrt{x^{2} y^{2}+3\left(x^{2}+y^{2}\right)+9} \geq 19-3 x y \\ \Leftrightarrow & 4-2 t+6+2 \sqrt{t^{2}+3(4-2 t)+9} \geq 19-3 t \\ \Leftrightarrow & 2 \sqrt{t^{2}-6 t+21} \geq 9-t \\ \Leftrightarrow & 4\left(t^{2}-6 t+21\right) \geq 81-18 t+t^{2} \end{aligned} \]

\[ \begin{aligned} & \Leftrightarrow 3 t^{2}-6 t+3 \geq 0 \\ & \Leftrightarrow t^{2}-2 t+1 \geq 0 \\ & \Leftrightarrow(t-1)^{2} \geq 0 \end{aligned} \] which is obvious. Equality holds true for \(t=1\), i.e. \(x y=1 \Rightarrow y=\tfrac{1}{x} \Rightarrow x+\tfrac{1}{x}=2 \Rightarrow(x-1)^{2}=0 \Rightarrow x=1\) and \(y=1\). Thus, the inequality (2) is proved.

\(2^{0}\). Now, we will prove the inequality (3), i.e.:

\[ \begin{aligned} & \sqrt{19-3 x y}+\sqrt{x y+3} \geq 6 \\ \Leftrightarrow & 19-3 x y+x y+3+2 \sqrt{19-3 x y} \cdot \sqrt{x y+3} \geq 36 \\ \Leftrightarrow & 2 \sqrt{10 x y+57-3 x^{2} y^{2}} \geq 14+2 x y \\ \Leftrightarrow & \sqrt{10 t+57-3 t^{2}} \geq 7+t \\ \Leftrightarrow & 10 t+57-3 t^{2} \geq 49+14 t+t^{2} \\ \Leftrightarrow & 4 t^{2}+4 t-8 \leq 0 \\ \Leftrightarrow & t^{2}+t-2 \leq 0 \\ \Leftrightarrow & (t-1)(t+2) \leq 0 \end{aligned} \] which is true because \(0 \leq t \leq 1\), i.e. \(t-1 \leq 0\). Equality holds, for \(t=1\), i.e. for \(x=y=1\). Thus, the inequality (3) is also proved.

The given inequality (1) follows from (2) and (3).

In the proof above a very important question arises: How to come to the inequality (2)? We will give an answer to this question in the proof of the next inequality:

(4) \[ \sqrt{x^{2}+8}+\sqrt{y^{2}+8}+\sqrt{x y+8} \geq 9 ;(x, y \geq 0, x+y=2) \]

Proof: We will consider the inequality:

(5) \[ \sqrt{x^{2}+8}+\sqrt{y^{2}+8} \geq \sqrt{\alpha-\beta x y} ; \quad(\alpha, \beta \in \mathbb{Q}) \]

(6) \[ \begin{aligned} & \Leftrightarrow x^{2}+8+y^{2}+8+2 \sqrt{x^{2}+8} \cdot \sqrt{y^{2}+8} \geq \alpha-\beta x y \\ & \Leftrightarrow x^{2}+y^{2}+16+2 \sqrt{x^{2} y^{2}+8\left(x^{2}+y^{2}\right)+64} \geq \alpha-\beta x y \\ & \Leftrightarrow 4-2 t+16+2 \sqrt{t^{2}+8(4-2 t)+64} \geq \alpha-\beta t \\ & \Leftrightarrow 2 \sqrt{t^{2}-16 t+96} \geq \alpha-20+(2-\beta) t \end{aligned} \]

Equality holds in (5) for \(x=y=1\), and we get from (5) that \(\sqrt{\alpha-\beta}=6\), i.e. \(\alpha-\beta=36\).

The inequality (6) is equivalent now with the inequality:

\[ \begin{gathered} 2 \sqrt{t^{2}-16 t+96} \geq 16+\beta+(2-\beta) t \\ \Leftrightarrow 4\left(t^{2}-16 t+96\right) \geq(16+\beta)^{2}+2(16+\beta)(2-\beta) t+(2-\beta)^{2} t^{2} \end{gathered} \]

Let now

\[ P(t)=\left[4-(2-\beta)^{2}\right] t^{2}+[-64-2(16+\beta)(2-\beta)] t+384-(16+\beta)^{2} \geq 0 \]

Since \(P(1)\) is the minimal value of the polynomial in the interval under consideration, then it follows that \(P^{\prime}(1)=0\), i.e.:

\[ \begin{aligned} & P^{\prime}(t)=2\left[4-(2-\beta)^{2}\right] t-64-2(16+\beta)(2-\beta) \\ & \Rightarrow P^{\prime}(1)=8-2(2-\beta)^{2}-64-2(16+\beta)(2-\beta)=0 \\ & \Rightarrow 4-\left(4-4 \beta+\beta^{2}\right)-32-\left(32-14 \beta-\beta^{2}\right)=0 \\ & \Rightarrow 4-4+4 \beta-\beta^{2}-32-32+14 \beta+\beta^{2}=0 \\ & \Rightarrow 18 \beta-64=0 \\ & \Rightarrow \beta=\tfrac{32}{9} \end{aligned} \]

and from here (because \(\alpha-\beta=36\) ) we get \(\alpha=\tfrac{356}{9}\).

We will now prove the inequality:

(7) \[ \sqrt{x^{2}+8}+\sqrt{y^{2}+8} \geq \sqrt{\tfrac{356}{9}-\tfrac{32}{9} x y} \]

\[ \begin{aligned} & \Leftrightarrow x^{2}+8+y^{2}+8+2 \sqrt{x^{2}+8} \cdot \sqrt{y^{2}+8} \geq \tfrac{356}{9}-\tfrac{32}{9} x y \\ & \Leftrightarrow 4-2 t+16+2 \sqrt{t^{2}-16 t+96} \geq \tfrac{356}{9}-\tfrac{32}{9} t \\ & \Leftrightarrow 9 \sqrt{t^{2}-16 t+96} \geq 88-7 t \\ & \Leftrightarrow 81\left(t^{2}-16 t+96\right) \geq 7744-1232 t+49 t^{2} \\ & \Leftrightarrow 32 t^{2}-64 t+32 \geq 0 \\ & \Leftrightarrow t^{2}-2 t+1 \geq 0 \\ & \Leftrightarrow(t-1)^{2} \geq 0 \end{aligned} \] which is obvious. Equality holds for \(t=1\), i.e. \(x=y=1\). Thus, the inequality (7) is proved.

Finally, we will prove the inequality:

(8) \[ \begin{aligned} & \qquad \sqrt{\tfrac{356}{9}-\tfrac{32}{9} x y}+\sqrt{x y+8} \geq 9 \\ & \Leftrightarrow \tfrac{356}{9}-\tfrac{32}{9} x y+x y+8+2 \sqrt{\tfrac{356}{9}-\tfrac{32}{9} x y} \cdot \sqrt{x y+8} \geq 81 \\ & \Leftrightarrow 356-32 x y+9 x y+72+6 \sqrt{(356-32 x y)(x y+8)} \geq 729 \\ & \Leftrightarrow 428-23 t+6 \sqrt{100 t-32 t^{2}+2848} \geq 729 \\ & \Leftrightarrow 6 \sqrt{100 t-32 t^{2}+2848} \geq 301+23 t \\ & \Leftrightarrow 36\left(100 t-32 t^{2}+2848\right) \geq 90601+529 t^{2}+13846 t \\ & \Leftrightarrow 1681 t^{2}+10246 t-11927 \leq 0 \\ & \Leftrightarrow(t-1)(1681 t+11927) \leq 0 \end{aligned} \]

which is true because \(0 \leq t \leq 1\), i.e. \(t-1 \leq 0\). Equality holds for \(t=1\), i.e. \(x=y=1\). Thus, the inequality (8) is proved.

Now, the given inequality (4) follows from (7) and (8).

We leave the following inequality to the reader:

(9) \[ \sqrt{x^{2}+15}+\sqrt{y^{2}+15}+\sqrt{x y+15} \geq 12 ;(x, y \geq 0, x+y=2) \]

For the proof of this inequality one could consider two other inequalities:

(10) \[ \sqrt{x^{2}+15}+\sqrt{y^{2}+15} \geq \sqrt{\alpha-\beta x y} ;(\alpha, \beta \in \mathbb{Q}) \]

and

(11) \[ \sqrt{\alpha-\beta x y}+\sqrt{x y+15} \geq 12 \]

Finally, we formulate a generalised version:

(12) \[ \sqrt{x^{2}+n^{2}-1}+\sqrt{y^{2}+n^{2}-1}+\sqrt{x y+n^{2}-1} \geq 3 n \]

where \(x, y \geq 0, x+y=2\) and \(n \in \mathbb{N},(n \geq 2)\).

Much work but at the same much pleasure. This is Mathematics.

NOTES/БЕЛЕЖКИ

1. AOPS-Art of ProblemSolving (web page).

REFERENCES/ЛИТЕРАТУРА

Arslanagić, Š. (2004). Matematika za nadarene. Sarajevo: Bosanska riječ.

Cirtoaje, V. (2006). Algebraic Inequalities – Old and New Methods. Zalau: Gil Publishing House.

Cvetkovski, Z. (2012). Inequalities – Theorems, Techniques and Selected Problems. Berlin Heidelberg: Springer Verlag.