x+y+z=4‌and‌x2+y2+z2=6 ∴y + z = 4 - x yz=‌‌
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{(y+z)2−(y2+z2)} =‌‌
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{(4−x)2−(6−x2)} ⇒‌‌yz=x2−4x+5 Hence, y and z are the roots of t2−(4−x)t+(x2−4x+5)≥0 Since the roots y and z are real (4−x)2−4(x2−4x+5)≥0 ⇒‌3x2−8x−4≤0 ⇒(3x−2)(x−2)≤0 ⇒x∈[‌‌