Isabelle Yeong
Isabelle's posts
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You can find ❤️ in ...... Math!
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Given the positive real numbers $a,\,b,\,c$ and $x,\,y,\,z$ satisfying the condition: $a+x=b+y=c+z=1$ Prove the inequality [MATH]\left(abc+xyz\right)\left(\frac{1}{ay}+\frac{1}{bz}+\frac{1}{cx}\right)\ge 3[/MATH].
Given the positive real numbers $a,\,b,\,c$ and $x,\,y,\,z$ satisfying the condition: $a+x=b+y=c+z=1$ Prove the inequality [MATH]\left(abc+xyz\right)\left(\frac{1}{ay}+\frac{1}{bz}+\frac{1}{cx}\right)\ge 3[/MATH]. My solution: Rewrite the intended LHS of th...﻿
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Evaluate [MATH]\small\left\lfloor{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)}\right\rfloor[/MATH] without using a calculator.
Evaluate [MATH]\left\lfloor{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)}\right\rfloor[/MATH] without using a calculator. My solution: [MATH]\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\l...﻿
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If $a,\,b$ and $c$ are the lengths of the sides of a right triangle with hypotenuse $c$, prove [MATH]\frac{(c − a)(c − b)}{(c + a)(c + b)}\le 17 − 12\sqrt{2}[/MATH].
Show that if $a,\,b$ and $c$ are the lengths of the sides of a right triangle with hypotenuse $c$, then (a) [MATH]\frac{c}{\sqrt{ab}}\ge \sqrt{2}[/MATH] (b) [MATH]\frac{(c − a)(c − b)}{(c + a)(c + b)}\le 17 − 12\sqrt{2}[/MATH]. My solution: (a) $\begin{alig...﻿ Post has attachment Solve for real solution of the system below: [MATH]\left\lfloor{x^3}\right\rfloor+\left\lfloor{x^2}\right\rfloor+\left\lfloor{x}\right\rfloor=\left\{x\right\}-1[/MATH]. Solve for real solution of the system below: [MATH]\left\lfloor{x^3}\right\rfloor+\left\lfloor{x^2}\right\rfloor+\left\lfloor{x}\right\rfloor=\left\{x\right\}-1[/MATH]. My solution: First, observe that the LHS of the equality must yield an integer, this tel...﻿ Post has attachment Solve for real solutions for the system:$x+y+z=-a,\\x^2+y^2+z^2=a^2,\\x^3+y^3+z^3=-a^3.$Solve for real solutions for the system:$x+y+z=-a,\\x^2+y^2+z^2=a^2,\\x^3+y^3+z^3=-a^3.$My solution: Let$x,\,y$and$z$be the real roots for a cubic polynomial. From the relation$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$, we have:$(-a)^2=a^2+2(xy+yz+zx)a^...﻿
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Show that [MATH]abc\le (ab+bc+ca)(a^2+b^2+c^2)^2[/MATH] for all positive real $a,\,b$ and $c$ such that $a+b+c=1.$
Show that [MATH]abc\le (ab+bc+ca)(a^2+b^2+c^2)^2[/MATH] for all positive real $a,\,b$ and $c$ such that $a+b+c=1.$ My solution: [MATH](ab+bc+ca)(a^2+b^2+c^2)^2[/MATH] [MATH]\ge \frac{(ab+bc+ca)(a^2+b^2+c^2)(a+b+c)^2}{3}[/MATH] since $3(a^2+b^2+c^2)\ge (a+b+...﻿ Post has attachment Find the real solution(s) to the system$\sqrt{4a-b^2}=\sqrt{b+2}+\sqrt{4a^2+b}$. Find the real solution(s) to the system$\sqrt{4a-b^2}=\sqrt{b+2}+\sqrt{4a^2+b}$. My solution: We know that for any two modulus functions, their sum is always greater than or equals to , we therefore have:$\sqrt{b+2}+\sqrt{4a^2+b}\ge \sqrt{b+2+4a^2+b}$But...﻿ Post has attachment Prove that [MATH]\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+b^3+abc}\le \frac{1}{abc}[/MATH] for all positive real$a,\,b$and$c$. Prove that [MATH]\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+b^3+abc}\le \frac{1}{abc}[/MATH] for all positive real$a,\,b$and$c$. My solution: [MATH]\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc}[/MATH] [MATH]=\frac{1}{(a+...﻿ Post has attachment Let$a,\,b,\,c,\,x,\,y$and$z$be strictly positive real numbers, prove that [MATH](a+x)(b+y)(c+z)+4\left(\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}\right)\ge 20[/MATH]. Let$a,\,b,\,c,\,x,\,y$and$z\$ be strictly positive real numbers, prove that [MATH](a+x)(b+y)(c+z)+4\left(\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}\right)\ge 20[/MATH]. My solution: [MATH](a+x)(b+y)(c+z)+4\left(\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}\right)[/M...﻿