1. Muammo: Agar $a > 0$ bo'lsa, quyidagi tengsizlikni isbotlang: $$\frac{a+9}{2} + \frac{a+16}{2} > 7\sqrt{a}$$ 2. Tengsizlikni soddalashtirish uchun chap tomonni jamlaymiz: $$\frac{a+9}{2} + \frac{a+16}{2} = \frac{(a+9) + (a+16)}{2} = \frac{2a + 25}{2} = a + \frac{25}{2}$$ 3. Tengsizlik endi quyidagicha ko'rinishda: $$a + \frac{25}{2} > 7\sqrt{a}$$ 4. Ikkala tomonini $2$ ga ko'paytiramiz: $$2a + 25 > 14\sqrt{a}$$ 5. $\sqrt{a} = x$ deb olamiz, bunda $x > 0$: $$2x^2 + 25 > 14x$$ 6. Tengsizlikni chap tomonga o'tkazamiz: $$2x^2 - 14x + 25 > 0$$ 7. Kvadrat tenglama ko'rinishida: $$2x^2 - 14x + 25 = 0$$ 8. Diskriminantni hisoblaymiz: $$D = (-14)^2 - 4 \cdot 2 \cdot 25 = 196 - 200 = -4 < 0$$ 9. Diskriminant manfiy, demak kvadrat ifoda hech qachon nolga teng emas va doimo musbat: $$2x^2 - 14x + 25 > 0 \quad \text{har doim}$$ 10. Shunday qilib, tengsizlik har doim to'g'ri, ya'ni isbotlandi. Javob: Tengsizlik $a > 0$ uchun har doim to'g'ri.