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Algebra Difficulty 7.0 National olympiad, round 2 Prove it

Example 5 Let a,b,x,y,ka, b, x, y, k be positive numbers, and k<2,a2+b2−kab=x2+y2−kxy=1k<2, a^{2}+b^{2}-k a b=x^{2}+y^{2}-k x y=1, prove: ∣ax−by∣⩽24−k2,∣ay+bx−kby∣⩽24−k2|a x-b y| \leqslant \frac{2}{\sqrt{4-k^{2}}},|a y+b x-k b y| \leqslant \frac{2}{\sqrt{4-k^{2}}}.

Solutions — 2

Solution 1

Prove that because a2+b2−kab=1a^{2}+b^{2}-k a b=1, so (a−kb2)2+(4−k22b)2=1\left(a-\frac{k b}{2}\right)^{2}+\left(\frac{\sqrt{4-k^{2}}}{2} b\right)^{2}=1, similarly (4−k22x)2+(kx2−y)2=1\left(\frac{\sqrt{4-k^{2}}}{2} x\right)^{2}+\left(\frac{k x}{2}-y\right)^{2}=1. Apply the Cauchy-Schwarz inequality:
((a−kb2)2+(4−k22b)2][(4−k22x)2+(kx2−y)2]⩾[(a−kb2)(4−k22x)+4−k22b(kx2−y)]2=[4−k22(ax−by)]2\begin{array}{l} \left(\left(a-\frac{k b}{2}\right)^{2}+\left(\frac{\sqrt{4-k^{2}}}{2} b\right)^{2}\right]\left[\left(\frac{\sqrt{4-k^{2}}}{2} x\right)^{2}+\left(\frac{k x}{2}-y\right)^{2}\right] \geqslant \\ {\left[\left(a-\frac{k b}{2}\right)\left(\frac{\sqrt{4-k^{2}}}{2} x\right)+\frac{\sqrt{4-k^{2}}}{2} b\left(\frac{k x}{2}-y\right)\right]^{2}=\left[\frac{\sqrt{4-k^{2}}}{2}(a x-b y)\right]^{2}} \end{array}

Thus,
∣ax−by∣⩽24−k2|a x-b y| \leqslant \frac{2}{\sqrt{4-k^{2}}}

By swapping the positions of xx and yy and making appropriate sign changes, noting that (a−kb2)2+(4−k22b)2=1\left(a-\frac{k b}{2}\right)^{2}+\left(\frac{\sqrt{4-k^{2}}}{2} b\right)^{2}=1 and (4−k22y)2+(x−ky2)2=1\left(\frac{\sqrt{4-k^{2}}}{2} y\right)^{2}+\left(x-\frac{k y}{2}\right)^{2}=1, apply the Cauchy-Schwarz inequality:
[(a−kb2)2+(4−k22b)2][(4−k22y)2+(x−ky2)2]⩾[(a−kb2)(4−k22y)+4−k22b(x−ky2)]2=[4−k22(ay+bx−kby)]2\begin{array}{l} {\left[\left(a-\frac{k b}{2}\right)^{2}+\left(\frac{\sqrt{4-k^{2}}}{2} b\right)^{2}\right]\left[\left(\frac{\sqrt{4-k^{2}}}{2} y\right)^{2}+\left(x-\frac{k y}{2}\right)^{2}\right] \geqslant} \\ {\left[\left(a-\frac{k b}{2}\right)\left(\frac{\sqrt{4-k^{2}}}{2} y\right)+\frac{\sqrt{4-k^{2}}}{2} b\left(x-\frac{k y}{2}\right)\right]^{2}=} \\ {\left[\frac{\sqrt{4-k^{2}}}{2}(a y+b x-k b y)\right]^{2}} \end{array}

Thus,
∣ay+bx−kby∣⩽24−k2|a y+b x-k b y| \leqslant \frac{2}{\sqrt{4-k^{2}}}

Solution 2

Prove that because a2+b2−kab=1a^{2}+b^{2}-k a b=1, so
(a−kb2)2+(4−k22b)2=1\left(a-\frac{k b}{2}\right)^{2}+\left(\frac{\sqrt{4-k^{2}}}{2} b\right)^{2}=1

Similarly, (4−k22x)2+(kx2−y)2=1\left(\frac{\sqrt{4-k^{2}}}{2} x\right)^{2}+\left(\frac{k x}{2}-y\right)^{2}=1.
Applying the Cauchy-Schwarz inequality, we have
[(a−kb2)2+(4−k22b)2]⋅[(4−k22x)2+(kx2−y)2]⩾[(a−kb2)(4−k22x)+4−k22b(kx2−y)]2=[4−k22(ax−by)]2,\begin{aligned} & {\left[\left(a-\frac{k b}{2}\right)^{2}+\left(\frac{\sqrt{4-k^{2}}}{2} b\right)^{2}\right] \cdot\left[\left(\frac{\sqrt{4-k^{2}}}{2} x\right)^{2}+\left(\frac{k x}{2}-y\right)^{2}\right] } \\ \geqslant & {\left[\left(a-\frac{k b}{2}\right)\left(\frac{\sqrt{4-k^{2}}}{2} x\right)+\frac{\sqrt{4-k^{2}}}{2} b\left(\frac{k x}{2}-y\right)\right]^{2} } \\ = & {\left[\frac{\sqrt{4-k^{2}}}{2}(a x-b y)\right]^{2}, } \end{aligned}

Thus, ∣ax−by∣⩽24−k2|a x-b y| \leqslant \frac{2}{\sqrt{4-k^{2}}}.
By swapping xx and yy and appropriately changing the sign, noting that (a−kb2)2+\left(a-\frac{k b}{2}\right)^{2}+ (4−k˙22b)2=1\left(\frac{\sqrt{4-\dot{k}^{2}}}{2} b\right)^{2}=1 and (4−k22y)2+(x−ky2)2=1\left(\frac{\sqrt{4-k^{2}}}{2} y\right)^{2}+\left(x-\frac{k y}{2}\right)^{2}=1, applying the Cauchy-Schwarz inequality again, we have
[(a−kb2)2+(4−k22b)2]⋅[(4−k22y)2+(x−ky2)2]⩾[(a−kb2)(4−k22y)+4−k22b(x−ky2)]2=[4−k22(ay+bx−kby)]2,\begin{aligned} & {\left[\left(a-\frac{k b}{2}\right)^{2}+\left(\frac{\sqrt{4-k^{2}}}{2} b\right)^{2}\right] \cdot\left[\left(\frac{\sqrt{4-k^{2}}}{2} y\right)^{2}+\left(x-\frac{k y}{2}\right)^{2}\right] } \\ \geqslant & {\left[\left(a-\frac{k b}{2}\right)\left(\frac{\sqrt{4-k^{2}}}{2} y\right)+\frac{\sqrt{4-k^{2}}}{2} b\left(x-\frac{k y}{2}\right)\right]^{2} } \\ = & {\left[\frac{\sqrt{4-k^{2}}}{2}(a y+b x-k b y)\right]^{2}, } \end{aligned}

Thus, ∣ay+bx−kby∣⩽24−k2|a y+b x-k b y| \leqslant \frac{2}{\sqrt{4-k^{2}}}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.